Explanation:
Okay, so I need to solve this problem about an unfair six-sided die. Let me try to break it down step by step.

First, the die has faces numbered 1 through 6. The key points given are:

1. The probability of obtaining face F is greater than 1/6.
2. The probability of obtaining the face opposite to F is less than 1/6.
3. The other four faces have probability 1/6 each.
4. The sum of numbers on each pair of opposite faces is 7. So, for example, 1 is opposite 6, 2 opposite 5, 3 opposite 4. That makes sense because 1+6=7, 2+5=7, 3+4=7.

Now, when two such dice are rolled, the probability of getting a sum of 7 is 47/288. We need to find the probability of face F, which is m/n, and then compute m + n.

Let me think. Let me denote the probability of face F as p. Then, the probability of its opposite face, let's call it F', is q. The other four faces each have probability 1/6. Since the die is six-sided, the total probability must add up to 1. So:

p + q + 4*(1/6) = 1

Simplify that:

p + q + 2/3 = 1 => p + q = 1 - 2/3 = 1/3.

So p + q = 1/3. Also, we know that p > 1/6 and q < 1/6.

Now, when two dice are rolled, the probability of getting a sum of 7 is 47/288. Let me think about how to compute that probability.

First, the possible ways to get a sum of 7 with two dice. Normally, with fair dice, there are 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). But here, the dice are unfair. So the probability of each pair depends on the probabilities of the individual faces.

But since the dice are identical in terms of their structure (each has opposite faces summing to 7, etc.), maybe we can model the probability accordingly.

But first, let me note that each die has opposite faces summing to 7. So, for each die, the pairs (1,6), (2,5), (3,4) are opposite. So, if F is one of these faces, then its opposite is the other face in the pair. Let me assume that F is one of the numbers 1, 2, 3, 4, 5, or 6. But since the problem says "face F" and "opposite face F", so F is a specific face, and its opposite is another specific face. Let me think about how the sum of 7 can be achieved.

When rolling two dice, the sum of 7 can be achieved in several ways:

- One die shows a number, and the other shows its opposite. For example, if one die shows 1, the other needs to show 6; if one die shows 2, the other needs to show 5, etc. But since the dice are unfair, the probabilities of these pairs will vary.

But also, there are other ways to get a sum of 7. For example, if one die shows 3 and the other shows 4, which are opposites. But since each die has opposite faces summing to 7, each pair of opposite faces is a pair that sums to 7. Therefore, the sum of 7 can be achieved by rolling a pair of opposite faces on the two dice. However, there's also the case where one die shows a number and the other shows a different number that isn't its opposite. Wait, no. Wait, if you have two numbers that are not opposites, their sum can be 7? Let me check.

For example, if one die shows 1 and the other shows 6, that's 7. If one die shows 2 and the other shows 5, that's 7. Similarly for 3 and 4. But if you have, say, 1 and 5, that's 6. 1 and 4 is 5. So the only way to get a sum of 7 is if the two numbers are opposites. Therefore, the sum of 7 can only be achieved by rolling a pair of opposite faces. Therefore, the total probability of getting a sum of 7 is the sum over all pairs of opposite faces of the probability that the first die shows one face and the second die shows its opposite.

But since the dice are identical, we need to consider all possible pairs. However, since each die has three pairs of opposite faces: (1,6), (2,5), (3,4). Let me think.

But here's the thing: the die has one face F with probability p, its opposite face F' with probability q, and the other four faces each with probability 1/6. So depending on which face F is, the pairs will be different. However, since the problem doesn't specify which face F is, but mentions that the sum of numbers on each pair of opposite faces is 7, so regardless of which face F is, the opposite face is fixed. For example, if F is 1, then F' is 6; if F is 2, F' is 5, etc.

But since the problem says "the probability of obtaining face F is greater than 1/6, the probability of obtaining the face opposite face F is less than 1/6", so F is one of the faces with higher probability, and its opposite is the one with lower probability.

But since the sum of 7 can be achieved by rolling any pair of opposite faces. So, for each of the three pairs (1,6), (2,5), (3,4), we can compute the probability of getting that pair. However, for the pair that includes F and F', their probabilities are p and q, respectively. For the other two pairs, each of their faces has probability 1/6. Let me think.

Let me formalize this. Let me suppose that F is one of the numbers, say, for example, F is 1. Then F' is 6. Then, the pairs that sum to 7 are (1,6), (2,5), (3,4). So, the probability of getting a sum of 7 is:

P(1 on first die and 6 on second die) + P(6 on first die and 1 on second die) + P(2 on first die and 5 on second die) + P(5 on first die and 2 on second die) + P(3 on first die and 4 on second die) + P(4 on first die and 3 on second die).

But since the dice are identical, the probability of (a, b) is P(a) * P(b), and since the dice are independent. So, for each pair (a, b) where a + b = 7, we have to compute P(a) * P(b) + P(b) * P(a) = 2 * P(a) * P(b) if a ≠ b. Wait, but for the pair (a, b) where a and b are different, but since the die is rolled twice, the order matters. However, since the problem says "the probability of obtaining a sum of 7", which is the same regardless of the order. So, for each pair of opposite faces, we have two ordered pairs: (a, b) and (b, a). Therefore, the total probability is the sum over all three pairs of 2 * P(a) * P(b). Except for the case where a = b? But since each pair of opposite faces are distinct numbers, so a ≠ b. Therefore, for each of the three pairs, we have two ordered pairs, so total of 6 ordered pairs, but since each unordered pair is counted twice. However, since the sum is 7, each unordered pair corresponds to two ordered pairs. Therefore, the total probability is:

For each of the three pairs (a, b) where a + b = 7, compute 2 * P(a) * P(b). Therefore, the total probability is:

2 * [P(1)P(6) + P(2)P(5) + P(3)P(4)].

But in our case, for the die, we have:

- For the pair (F, F'), which is one of the three pairs, say (F, F'), their probabilities are p and q.

- For the other two pairs, each of their faces has probability 1/6. For example, if F is 1 and F' is 6, then the other pairs are (2,5) and (3,4). Each of these pairs has both faces with probability 1/6. Therefore, for each of these pairs, P(a) * P(b) = (1/6)(1/6) = 1/36. Therefore, the total probability would be:

2 * [p*q + 1/36 + 1/36] = 2 * [pq + 2/36] = 2 * [pq + 1/18] = 2pq + 1/9.

But the problem says that the probability of getting a sum of 7 is 47/288. Therefore:

2pq + 1/9 = 47/288.

Let me check that again. Wait, if the three pairs are (F, F'), (a, b), (c, d), then the total probability is 2*(p*q + (1/6)*(1/6) + (1/6)*(1/6)) = 2*(pq + 1/36 + 1/36) = 2pq + 2*(1/36)*2? Wait, no. Wait, the three pairs contribute:

For the first pair (F, F'): 2 * p * q.

For the second pair (a, b): 2 * (1/6) * (1/6) = 2*(1/36) = 1/18.

Similarly for the third pair (c, d): 2*(1/6)*(1/6) = 1/18.

Therefore, total probability is 2pq + 1/18 + 1/18 = 2pq + 1/9. So that's correct.

So, 2pq + 1/9 = 47/288.

Let me compute 1/9 as 32/288. Therefore:

2pq + 32/288 = 47/288 => 2pq = 47/288 - 32/288 = 15/288 = 5/96.

Therefore, pq = 5/192.

So, we have two equations:

1. p + q = 1/3.

2. pq = 5/192.

So, we can solve for p and q. Let me think. Let me denote p and q as roots of the quadratic equation x^2 - (p + q)x + pq = 0 => x^2 - (1/3)x + 5/192 = 0.

Let me compute the discriminant:

D = (1/3)^2 - 4*(5/192) = 1/9 - 20/192.

Convert to common denominator. Let me compute 1/9 as 21.333.../192? Wait, better to convert both fractions to denominator 192.

1/9 = 192/9 / 192? Wait, no. Let me compute:

Multiply numerator and denominator by 192/192:

Wait, 1/9 = (1 * 21.333...)/192. Wait, maybe better to compute decimal or find common denominator.

Alternatively, compute 1/9 - 20/192:

First, note that 1/9 = 64/576, and 20/192 = 60/576. Therefore, 64/576 - 60/576 = 4/576 = 1/144. Therefore, discriminant D = 1/144.

Therefore, solutions are:

x = [1/3 ± sqrt(1/144)] / 2 = [1/3 ± 1/12]/2.

Compute:

First, 1/3 is 4/12, so:

[4/12 ± 1/12]/2 = [ (4 ± 1)/12 ] / 2 = (5/12)/2 or (3/12)/2 = 5/24 or 1/8.

Therefore, the solutions are p = 5/24 and q = 1/8, or p = 1/8 and q = 5/24.

But we know that p > 1/6 and q < 1/6. Let me check:

1/6 is approximately 0.1667.

5/24 is approximately 0.2083, which is greater than 1/6. 1/8 is 0.125, which is less than 1/6. Therefore, since p > 1/6, p must be 5/24, and q is 1/8.

Therefore, the probability of face F is 5/24. Therefore, m = 5, n = 24. Are they coprime? 5 and 24 share no common factors besides 1. So yes. Therefore, m + n = 5 + 24 = 29.

Wait, but let me check again. Let me verify the calculations step by step to make sure I didn't make a mistake.

First, the total probability of sum 7 is 2pq + 1/9 = 47/288. Then, solving for pq:

2pq = 47/288 - 1/9 = 47/288 - 32/288 = 15/288 = 5/96. Therefore, pq = 5/192. Then, solving p + q = 1/3 and pq = 5/192. Then quadratic equation x^2 - (1/3)x + 5/192 = 0. Then discriminant D = (1/3)^2 - 4*(5/192) = 1/9 - 20/192. Converting 1/9 to 21.333.../192? Wait, let me do fractions properly.

Convert 1/9 to denominator 192: 1/9 = (192/9)/192? Wait, no. Let me find a common denominator between 9 and 192. The least common multiple of 9 and 192. Let's see, 192 factors into 2^6 * 3, and 9 is 3^2. So LCM is 2^6 * 3^2 = 64 * 9 = 576. Therefore, convert both fractions to denominator 576:

1/9 = 64/576.

20/192 = (20 * 3)/576 = 60/576.

Therefore, D = 64/576 - 60/576 = 4/576 = 1/144. Correct. Then sqrt(1/144) = 1/12. Then solutions:

[1/3 ± 1/12]/2. Convert 1/3 to 4/12, so:

(4/12 + 1/12)/2 = 5/12 / 2 = 5/24.

(4/12 - 1/12)/2 = 3/12 / 2 = 1/8. Correct. So p = 5/24 and q = 1/8. Since p > 1/6 (which is 4/24), 5/24 is indeed greater. And q = 1/8 = 3/24, which is less than 4/24. So that's correct.

Therefore, the probability of face F is 5/24, so m + n is 5 + 24 = 29. Therefore, the answer is 29. Wait, but let me check once again if my initial assumption about the sum of 7 being 2pq + 1/9 is correct.

Let me think again. Suppose that the die has three pairs of opposite faces. For each pair, the probability of getting that pair in two dice is 2 * P(a) * P(b). For the pair (F, F'), which has probabilities p and q, so 2pq. For the other two pairs, each has both faces with probability 1/6, so each contributes 2*(1/6)*(1/6) = 2/36 = 1/18. Therefore, total is 2pq + 1/18 + 1/18 = 2pq + 1/9. Yes, that's correct. Then 2pq + 1/9 = 47/288. Then solving gives pq = 5/192. Then solving quadratic gives p = 5/24 and q = 1/8. So that seems correct.

But let me check with actual numbers. Let me compute 2pq + 1/9 with p = 5/24 and q = 1/8.

First, pq = (5/24)(1/8) = 5/192. Then 2pq = 10/192 = 5/96. Then 1/9 is 32/288. Let me convert 5/96 to 288 denominator: 5/96 = (5 * 3)/288 = 15/288. Then 15/288 + 32/288 = 47/288. Which matches the given probability. So that's correct. Therefore, the answer is 5/24, so m + n = 29.

But wait, the problem says "the probability of obtaining face F is m/n", so 5/24 is reduced, so m + n is 29. Therefore, the answer is 29. But let me check again if there's any other possible case. For example, what if F is not part of the pair (1,6), but another pair? Wait, but regardless of which pair F is in, the calculation would be the same. Because the other two pairs would still have probabilities 1/6 each. So whether F is in (1,6), (2,5), or (3,4), the calculation remains the same. Therefore, the answer is 29.

But let me check if there's a possibility that the sum of 7 can be achieved in other ways. Wait, for example, if you have two dice, and you have numbers that are not opposite but still sum to 7? Wait, for example, if one die shows 1 and the other shows 6, which are opposites. If one die shows 2 and the other shows 5, which are opposites. Similarly for 3 and 4. But if you have, say, one die shows 1 and the other shows 2, that's 3, not 7. Similarly, 1 and 3 is 4, etc. So the only way to get 7 is to have opposite faces. Therefore, the calculation is correct. Therefore, the answer is 29.

But just to be thorough, let me think again. Suppose that F is 1, so F' is 6. Then, the probability of getting a sum of 7 is:

P(1 on first die and 6 on second die) + P(6 on first die and 1 on second die) + P(2 on first die and 5 on second die) + P(5 on first die and 2 on second die) + P(3 on first die and 4 on second die) + P(4 on first die and 3 on second die).

Which is:

2 * P(1) * P(6) + 2 * P(2) * P(5) + 2 * P(3) * P(4).

Since P(1) = p, P(6) = q, and P(2) = P(5) = P(3) = P(4) = 1/6. Therefore:

2 * p * q + 2 * (1/6 * 1/6) + 2 * (1/6 * 1/6) = 2pq + 2*(1/36) + 2*(1/36) = 2pq + 4/36 = 2pq + 1/9. Which is the same as before. Therefore, correct.

Therefore, the answer is 29. Therefore, I think that is correct.

**Final Answer**
\boxed{29}

To solve the problem, we analyze the structure of the unfair six-sided die and the given probability conditions.

---

### **Understanding the Die Structure**

- The die has faces numbered 1 through 6.
- Each pair of opposite faces sums to 7:
  - (1, 6), (2, 5), (3, 4)
- One face, say $ F $, has a probability $ p > \frac{1}{6} $.
- The face opposite to $ F $, say $ F' $, has a probability $ q < \frac{1}{6} $.
- The remaining four faces each have probability $ \frac{1}{6} $.

Since the total probability must sum to 1, we have:

$$
p + q + 4 \cdot \frac{1}{6} = 1 \Rightarrow p + q = \frac{1}{3}
$$

---

### **Calculating the Probability of Rolling a Sum of 7**

When rolling two such dice, the only way to get a sum of 7 is by rolling a pair of opposite faces. There are three such pairs: (1,6), (2,5), (3,4).

Let’s compute the total probability of rolling a sum of 7:

- For the pair $ (F, F') $, the probability is $ 2pq $ (since both $ (F, F') $ and $ (F', F) $ are valid ordered pairs).
- For the other two pairs (e.g., (2,5) and (3,4)), each face has probability $ \frac{1}{6} $, so the contribution is $ 2 \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{18} $ for each pair.

Thus, the total probability of rolling a sum of 7 is:

$$
2pq + \frac{1}{18} + \frac{1}{18} = 2pq + \frac{1}{9}
$$

We are told this probability is $ \frac{47}{288} $, so:

$$
2pq + \frac{1}{9} = \frac{47}{288}
$$

Convert $ \frac{1}{9} $ to have denominator 288:

$$
\frac{1}{9} = \frac{32}{288}
$$

Now solve for $ pq $:

$$
2pq = \frac{47}{288} - \frac{32}{288} = \frac{15}{288} = \frac{5}{96}
\Rightarrow pq = \frac{5}{192}
$$

---

### **Solving the System of Equations**

We now have two equations:

1. $ p + q = \frac{1}{3} $
2. $ pq = \frac{5}{192} $

These are the sum and product of the roots of a quadratic equation:

$$
x^2 - \left( \frac{1}{3} \right)x + \frac{5}{192} = 0
$$

Multiply through by 192 to eliminate denominators:

$$
192x^2 - 64x + 5 = 0
$$

Use the quadratic formula:

$$
x = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 192 \cdot 5}}{2 \cdot 192}
= \frac{64 \pm \sqrt{4096 - 3840}}{384}
= \frac{64 \pm \sqrt{256}}{384}
= \frac{64 \pm 16}{384}
$$

So the two roots are:

$$
x = \frac{80}{384} = \frac{5}{24}, \quad x = \frac{48}{384} = \frac{1}{8}
$$

Since $ p > \frac{1}{6} $ and $ q < \frac{1}{6} $, we assign:

- $ p = \frac{5}{24} $
- $ q = \frac{1}{8} $

---

### **Final Step: Compute $ m + n $**

The probability of face $ F $ is $ \frac{5}{24} $, so $ m = 5 $, $ n = 24 $, and:

$$
m + n = 5 + 24 = \boxed{29}
$$

[Assistant]
Explanation:
Okay, so I need to solve this problem about an unfair six-sided die. Let me try to break it down step by step.

First, the die has faces numbered 1 through 6. The key points given are:

1. The probability of obtaining face F is greater than 1/6.
2. The probability of obtaining the face opposite to F is less than 1/6.
3. The other four faces have probability 1/6 each.
4. The sum of numbers on each pair of opposite faces is 7. So, for example, 1 is opposite 6, 2 opposite 5, 3 opposite 4. That makes sense because 1+6=7, 2+5=7, 3+4=7.

Now, when two such dice are rolled, the probability of getting a sum of 7 is 47/288. We need to find the probability of face F, which is m/n, and then compute m + n.

Let me think. Let me denote the probability of face F as p. Then, the probability of its opposite face, let's call it F', is q. The other four faces each have probability 1/6. Since the die is six-sided, the total probability must add up to 1. So:

p + q + 4*(1/6) = 1

Simplify that:

p + q + 2/3 = 1 => p + q = 1 - 2/3 = 1/3.

So p + q = 1/3. Also, we know that p > 1/6 and q < 1/6.

Now, when two dice are rolled, the probability of getting a sum of 7 is 47/288. Let me think about how to compute that probability.

First, the possible ways to get a sum of 7 with two dice. Normally, with fair dice, there are 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). But here, the dice are unfair. So the probability of each pair depends on the probabilities of the individual faces.

But since the dice are identical in terms of their structure (each has opposite faces summing to 7, etc.), maybe we can model the probability accordingly.

But first, let me note that each die has opposite faces summing to 7. So, for each die, the pairs (1,6), (2,5), (3,4) are opposite. So, if F is one of these faces, then its opposite is the other face in the pair. Let me assume that F is one of the numbers 1, 2, 3, 4, 5, or 6. But since the problem says "face F" and "opposite face F", so F is a specific face, and its opposite is another specific face. Let me think about how the sum of 7 can be achieved.

When rolling two dice, the sum of 7 can be achieved in several ways:

- One die shows a number, and the other shows its opposite. For example, if one die shows 1, the other needs to show 6; if one die shows 2, the other needs to show 5, etc. But since the dice are unfair, the probabilities of these pairs will vary.

But also, there are other ways to get a sum of 7. For example, if one die shows 3 and the other shows 4, which are opposites. But since each die has opposite faces summing to 7, each pair of opposite faces is a pair that sums to 7. Therefore, the sum of 7 can be achieved by rolling a pair of opposite faces on the two dice. However, there's also the case where one die shows a number and the other shows a different number that isn't its opposite. Wait, no. Wait, if you have two numbers that are not opposites, their sum can be 7? Let me check.

For example, if one die shows 1 and the other shows 6, that's 7. If one die shows 2 and the other shows 5, that's 7. Similarly for 3 and 4. But if you have, say, 1 and 5, that's 6. 1 and 4 is 5. So the only way to get a sum of 7 is if the two numbers are opposites. Therefore, the sum of 7 can only be achieved by rolling a pair of opposite faces. Therefore, the total probability of getting a sum of 7 is the sum over all pairs of opposite faces of the probability that the first die shows one face and the second die shows its opposite.

But since the dice are identical, we need to consider all possible pairs. However, since each die has three pairs of opposite faces: (1,6), (2,5), (3,4). Let me think.

But here's the thing: the die has one face F with probability p, its opposite face F' with probability q, and the other four faces each with probability 1/6. So depending on which face F is, the pairs will be different. However, since the problem doesn't specify which face F is, but mentions that the sum of numbers on each pair of opposite faces is 7, so regardless of which face F is, the opposite face is fixed. For example, if F is 1, then F' is 6; if F is 2, F' is 5, etc.

But since the problem says "the probability of obtaining face F is greater than 1/6, the probability of obtaining the face opposite face F is less than 1/6", so F is one of the faces with higher probability, and its opposite is the one with lower probability.

But since the sum of 7 can be achieved by rolling any pair of opposite faces. So, for each of the three pairs (1,6), (2,5), (3,4), we can compute the probability of getting that pair. However, for the pair that includes F and F', their probabilities are p and q, respectively. For the other two pairs, each of their faces has probability 1/6. Let me think.

Let me formalize this. Let me suppose that F is one of the numbers, say, for example, F is 1. Then F' is 6. Then, the pairs that sum to 7 are (1,6), (2,5), (3,4). So, the probability of getting a sum of 7 is:

P(1 on first die and 6 on second die) + P(6 on first die and 1 on second die) + P(2 on first die and 5 on second die) + P(5 on first die and 2 on second die) + P(3 on first die and 4 on second die) + P(4 on first die and 3 on second die).

But since the dice are identical, the probability of (a, b) is P(a) * P(b), and since the dice are independent. So, for each pair (a, b) where a + b = 7, we have to compute P(a) * P(b) + P(b) * P(a) = 2 * P(a) * P(b) if a ≠ b. Wait, but for the pair (a, b) where a and b are different, but since the die is rolled twice, the order matters. However, since the problem says "the probability of obtaining a sum of 7", which is the same regardless of the order. So, for each pair of opposite faces, we have two ordered pairs: (a, b) and (b, a). Therefore, the total probability is the sum over all three pairs of 2 * P(a) * P(b). Except for the case where a = b? But since each pair of opposite faces are distinct numbers, so a ≠ b. Therefore, for each of the three pairs, we have two ordered pairs, so total of 6 ordered pairs, but since each unordered pair is counted twice. However, since the sum is 7, each unordered pair corresponds to two ordered pairs. Therefore, the total probability is:

For each of the three pairs (a, b) where a + b = 7, compute 2 * P(a) * P(b). Therefore, the total probability is:

2 * [P(1)P(6) + P(2)P(5) + P(3)P(4)].

But in our case, for the die, we have:

- For the pair (F, F'), which is one of the three pairs, say (F, F'), their probabilities are p and q.

- For the other two pairs, each of their faces has probability 1/6. For example, if F is 1 and F' is 6, then the other pairs are (2,5) and (3,4). Each of these pairs has both faces with probability 1/6. Therefore, for each of these pairs, P(a) * P(b) = (1/6)(1/6) = 1/36. Therefore, the total probability would be:

2 * [p*q + 1/36 + 1/36] = 2 * [pq + 2/36] = 2 * [pq + 1/18] = 2pq + 1/9.

But the problem says that the probability of getting a sum of 7 is 47/288. Therefore:

2pq + 1/9 = 47/288.

Let me check that again. Wait, if the three pairs are (F, F'), (a, b), (c, d), then the total probability is 2*(p*q + (1/6)*(1/6) + (1/6)*(1/6)) = 2*(pq + 1/36 + 1/36) = 2pq + 2*(1/36)*2? Wait, no. Wait, the three pairs contribute:

For the first pair (F, F'): 2 * p * q.

For the second pair (a, b): 2 * (1/6) * (1/6) = 2*(1/36) = 1/18.

Similarly for the third pair (c, d): 2*(1/6)*(1/6) = 1/18.

Therefore, total probability is 2pq + 1/18 + 1/18 = 2pq + 1/9. So that's correct.

So, 2pq + 1/9 = 47/288.

Let me compute 1/9 as 32/288. Therefore:

2pq + 32/288 = 47/288 => 2pq = 47/288 - 32/288 = 15/288 = 5/96.

Therefore, pq = 5/192.

So, we have two equations:

1. p + q = 1/3.

2. pq = 5/192.

So, we can solve for p and q. Let me think. Let me denote p and q as roots of the quadratic equation x^2 - (p + q)x + pq = 0 => x^2 - (1/3)x + 5/192 = 0.

Let me compute the discriminant:

D = (1/3)^2 - 4*(5/192) = 1/9 - 20/192.

Convert to common denominator. Let me compute 1/9 as 21.333.../192? Wait, better to convert both fractions to denominator 192.

1/9 = 192/9 / 192? Wait, no. Let me compute:

Multiply numerator and denominator by 192/192:

Wait, 1/9 = (1 * 21.333...)/192. Wait, maybe better to compute decimal or find common denominator.

Alternatively, compute 1/9 - 20/192:

First, note that 1/9 = 64/576, and 20/192 = 60/576. Therefore, 64/576 - 60/576 = 4/576 = 1/144. Therefore, discriminant D = 1/144.

Therefore, solutions are:

x = [1/3 ± sqrt(1/144)] / 2 = [1/3 ± 1/12]/2.

Compute:

First, 1/3 is 4/12, so:

[4/12 ± 1/12]/2 = [ (4 ± 1)/12 ] / 2 = (5/12)/2 or (3/12)/2 = 5/24 or 1/8.

Therefore, the solutions are p = 5/24 and q = 1/8, or p = 1/8 and q = 5/24.

But we know that p > 1/6 and q < 1/6. Let me check:

1/6 is approximately 0.1667.

5/24 is approximately 0.2083, which is greater than 1/6. 1/8 is 0.125, which is less than 1/6. Therefore, since p > 1/6, p must be 5/24, and q is 1/8.

Therefore, the probability of face F is 5/24. Therefore, m = 5, n = 24. Are they coprime? 5 and 24 share no common factors besides 1. So yes. Therefore, m + n = 5 + 24 = 29.

Wait, but let me check again. Let me verify the calculations step by step to make sure I didn't make a mistake.

First, the total probability of sum 7 is 2pq + 1/9 = 47/288. Then, solving for pq:

2pq = 47/288 - 1/9 = 47/288 - 32/288 = 15/288 = 5/96. Therefore, pq = 5/192. Then, solving p + q = 1/3 and pq = 5/192. Then quadratic equation x^2 - (1/3)x + 5/192 = 0. Then discriminant D = (1/3)^2 - 4*(5/192) = 1/9 - 20/192. Converting 1/9 to 21.333.../192? Wait, let me do fractions properly.

Convert 1/9 to denominator 192: 1/9 = (192/9)/192? Wait, no. Let me find a common denominator between 9 and 192. The least common multiple of 9 and 192. Let's see, 192 factors into 2^6 * 3, and 9 is 3^2. So LCM is 2^6 * 3^2 = 64 * 9 = 576. Therefore, convert both fractions to denominator 576:

1/9 = 64/576.

20/192 = (20 * 3)/576 = 60/576.

Therefore, D = 64/576 - 60/576 = 4/576 = 1/144. Correct. Then sqrt(1/144) = 1/12. Then solutions:

[1/3 ± 1/12]/2. Convert 1/3 to 4/12, so:

(4/12 + 1/12)/2 = 5/12 / 2 = 5/24.

(4/12 - 1/12)/2 = 3/12 / 2 = 1/8. Correct. So p = 5/24 and q = 1/8. Since p > 1/6 (which is 4/24), 5/24 is indeed greater. And q = 1/8 = 3/24, which is less than 4/24. So that's correct.

Therefore, the probability of face F is 5/24, so m + n is 5 + 24 = 29. Therefore, the answer is 29. Wait, but let me check once again if my initial assumption about the sum of 7 being 2pq + 1/9 is correct.

Let me think again. Suppose that the die has three pairs of opposite faces. For each pair, the probability of getting that pair in two dice is 2 * P(a) * P(b). For the pair (F, F'), which has probabilities p and q, so 2pq. For the other two pairs, each has both faces with probability 1/6, so each contributes 2*(1/6)*(1/6) = 2/36 = 1/18. Therefore, total is 2pq + 1/18 + 1/18 = 2pq + 1/9. Yes, that's correct. Then 2pq + 1/9 = 47/288. Then solving gives pq = 5/192. Then solving quadratic gives p = 5/24 and q = 1/8. So that seems correct.

But let me check with actual numbers. Let me compute 2pq + 1/9 with p = 5/24 and q = 1/8.

First, pq = (5/24)(1/8) = 5/192. Then 2pq = 10/192 = 5/96. Then 1/9 is 32/288. Let me convert 5/96 to 288 denominator: 5/96 = (5 * 3)/288 = 15/288. Then 15/288 + 32/288 = 47/288. Which matches the given probability. So that's correct. Therefore, the answer is 5/24, so m + n = 29.

But wait, the problem says "the probability of obtaining face F is m/n", so 5/24 is reduced, so m + n is 29. Therefore, the answer is 29. But let me check again if there's any other possible case. For example, what if F is not part of the pair (1,6), but another pair? Wait, but regardless of which pair F is in, the calculation would be the same. Because the other two pairs would still have probabilities 1/6 each. So whether F is in (1,6), (2,5), or (3,4), the calculation remains the same. Therefore, the answer is 29.

But let me check if there's a possibility that the sum of 7 can be achieved in other ways. Wait, for example, if you have two dice, and you have numbers that are not opposite but still sum to 7? Wait, for example, if one die shows 1 and the other shows 6, which are opposites. If one die shows 2 and the other shows 5, which are opposites. Similarly for 3 and 4. But if you have, say, one die shows 1 and the other shows 2, that's 3, not 7. Similarly, 1 and 3 is 4, etc. So the only way to get 7 is to have opposite faces. Therefore, the calculation is correct. Therefore, the answer is 29.

But just to be thorough, let me think again. Suppose that F is 1, so F' is 6. Then, the probability of getting a sum of 7 is:

P(1 on first die and 6 on second die) + P(6 on first die and 1 on second die) + P(2 on first die and 5 on second die) + P(5 on first die and 2 on second die) + P(3 on first die and 4 on second die) + P(4 on first die and 3 on second die).

Which is:

2 * P(1) * P(6) + 2 * P(2) * P(5) + 2 * P(3) * P(4).

Since P(1) = p, P(6) = q, and P(2) = P(5) = P(3) = P(4) = 1/6. Therefore:

2 * p * q + 2 * (1/6 * 1/6) + 2 * (1/6 * 1/6) = 2pq + 2*(1/36) + 2*(1/36) = 2pq + 4/36 = 2pq + 1/9. Which is the same as before. Therefore, correct.

Therefore, the answer is 29. Therefore, I think that is correct.

**Final Answer**
\boxed{29}

To solve the problem, we analyze the structure of the unfair six-sided die and the given probability conditions.

---

### **Understanding the Die Structure**

- The die has faces numbered 1 through 6.
- Each pair of opposite faces sums to 7:
  - (1, 6), (2, 5), (3, 4)
- One face, say $ F $, has a probability $ p > \frac{1}{6} $.
- The face opposite to $ F $, say $ F' $, has a probability $ q < \frac{1}{6} $.
- The remaining four faces each have probability $ \frac{1}{6} $.

Since the total probability must sum to 1, we have:

$$
p + q + 4 \cdot \frac{1}{6} = 1 \Rightarrow p + q = \frac{1}{3}
$$

---

### **Calculating the Probability of Rolling a Sum of 7**

When rolling two such dice, the only way to get a sum of 7 is by rolling a pair of opposite faces. There are three such pairs: (1,6), (2,5), (3,4).

Let’s compute the total probability of rolling a sum of 7:

- For the pair $ (F, F') $, the probability is $ 2pq $ (since both $ (F, F') $ and $ (F', F) $ are valid ordered pairs).
- For the other two pairs (e.g., (2,5) and (3,4)), each face has probability $ \frac{1}{6} $, so the contribution is $ 2 \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{18} $ for each pair.

Thus, the total probability of rolling a sum of 7 is:

$$
2pq + \frac{1}{18} + \frac{1}{18} = 2pq + \frac{1}{9}
$$

We are told this probability is $ \frac{47}{288} $, so:

$$
2pq + \frac{1}{9} = \frac{47}{288}
$$

Convert $ \frac{1}{9} $ to have denominator 288:

$$
\frac{1}{9} = \frac{32}{288}
$$

Now solve for $ pq $:

$$
2pq = \frac{47}{288} - \frac{32}{288} = \frac{15}{288} = \frac{5}{96}
\Rightarrow pq = \frac{5}{192}
$$

---

### **Solving the System of Equations**

We now have two equations:

1. $ p + q = \frac{1}{3} $
2. $ pq = \frac{5}{192} $

These are the sum and product of the roots of a quadratic equation:

$$
x^2 - \left( \frac{1}{3} \right)x + \frac{5}{192} = 0
$$

Multiply through by 192 to eliminate denominators:

$$
192x^2 - 64x + 5 = 0
$$

Use the quadratic formula:

$$
x = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 192 \cdot 5}}{2 \cdot 192}
= \frac{64 \pm \sqrt{4096 - 3840}}{384}
= \frac{64 \pm \sqrt{256}}{384}
= \frac{64 \pm 16}{384}
$$

So the two roots are:

$$
x = \frac{80}{384} = \frac{5}{24}, \quad x = \frac{48}{384} = \frac{1}{8}
$$

Since $ p > \frac{1}{6} $ and $ q < \frac{1}{6} $, we assign:

- $ p = \frac{5}{24} $
- $ q = \frac{1}{8} $

---

### **Final Step: Compute $ m + n $**

The probability of face $ F $ is $ \frac{5}{24} $, so $ m = 5 $, $ n = 24 $, and:

$$
m + n = 5 + 24 = \boxed{29}
$$

Solution:
Let $p(a,b)$ denote the probability of obtaining  $a$ on the first die and $b$ on the second.  Then the probability of obtaining a sum of 7 is $$p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).$$Let the probability of obtaining face $F$ be $(1/6)+x$.  Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$. Therefore $$\begin{aligned}{{47}\over{288}}&=
4\left({1\over6}\right)^2+2\left({1\over6}+x\right)
\left({1\over6}-x\right)\cr&=
{4\over36}+2\left({1\over36}-x^2\right)\cr&=
{1\over6}-2x^2.\end{aligned}$$Then $2x^2=1/288$, and so $x=1/24$. The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$, and $m+n=\boxed{29}$.


Explanation:
Okay, so I need to find the sum of all complex solutions to this equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Hmm, complex solutions... So, I need to consider all roots of this equation, even if they are complex numbers. The question is asking for their sum.

First, let me think about how to approach this equation. It has rational expressions on the left side and a linear term on the right. The denominators are all quadratic in terms of $ x^2 $, so maybe I can simplify the equation by substituting $ y = x^2 $. Let me try that.

Let me set $ y = x^2 $. Then the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

Wait, but there's still an $ x $ term on the right side. If I substitute $ y = x^2 $, then $ x $ is still present. Maybe I need another substitution or approach. Alternatively, perhaps I can consider that the left side is a function of $ y $, and the right side is a function of $ x $. But since $ y = x^2 $, maybe I can express everything in terms of $ y $, but the problem is that the right side has $ x $, not $ y $.

Alternatively, maybe I can rearrange the equation to bring all terms to one side and then find a common denominator? But that might get complicated. Let me think.

Alternatively, maybe I can consider that if I have an equation involving $ x $, and I need to find the sum of all complex solutions. For polynomial equations, the sum of roots can be found using Vieta's formula. However, this equation isn't a polynomial equation because of the denominators. But perhaps if I can manipulate it into a polynomial equation, then Vieta's formula can be applied.

So, the plan is: first, eliminate denominators by multiplying both sides by the common denominator, which would be $ (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. However, this would result in a high-degree polynomial equation. Let me check the degrees.

Each denominator is quadratic, so multiplying them would give a degree 8 polynomial. However, the right side is $ 2010x - 4 $, so when multiplied by the common denominator, the right side would be $ (2010x - 4)(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. That would be a degree 9 polynomial. Therefore, the entire equation would be a degree 9 polynomial equation. However, the left side, when multiplied by the common denominator, would be a polynomial of degree 8. Therefore, moving everything to one side would result in a degree 9 polynomial.

But solving a degree 9 polynomial is not straightforward. However, the question is not asking for the roots, but the sum of all complex solutions. Maybe there's a trick here.

Wait, Vieta's formula tells us that the sum of the roots of a polynomial equation $ a_n x^n + \dots + a_0 = 0 $ is $ -a_{n-1}/a_n $. So, if I can write the equation as a polynomial equation, then I can find the sum of the roots. However, the problem is that the original equation has denominators, so when we multiply through by the common denominator, we might introduce extraneous roots (the ones that make denominators zero), but since we are considering complex solutions, maybe we need to be careful. However, the original equation is undefined when $ x^2 = 1, 2, 3, 4 $, so those values of $ x $ are excluded. Therefore, when we multiply both sides by the common denominator, we have to ensure that we don't include those roots. However, since we are looking for the sum of all complex solutions, which are the roots of the polynomial equation excluding those that make denominators zero. But maybe the sum of all roots of the polynomial equation (including those excluded) minus the sum of the excluded roots would give the desired sum. However, this seems complicated.

Alternatively, maybe there's a symmetry or substitution that can be used here. Let me think again.

Looking back at the original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Note that the left side is a function of $ x^2 $, while the right side is a linear function of $ x $. If I consider that the left side is even in $ x $, because replacing $ x $ with $ -x $ doesn't change $ x^2 $, so the left side is even. However, the right side is $ 2010x - 4 $, which is an odd function plus a constant. Therefore, the entire equation is not symmetric. However, if we consider that for each solution $ x $, there might be a corresponding solution $ -x $, but only if the right side satisfies some condition. Let me check.

Suppose $ x $ is a solution. Then, substituting $ -x $ into the equation:

Left side becomes the same as for $ x $, since $ (-x)^2 = x^2 $. The right side becomes $ 2010(-x) - 4 = -2010x - 4 $. Therefore, unless $ 2010x - 4 = -2010x - 4 $, which would imply $ 4020x = 0 $, i.e., $ x = 0 $, the equation is not symmetric. Therefore, except for $ x = 0 $, solutions don't come in pairs $ x $ and $ -x $. However, if $ x = 0 $ is a solution, then it would satisfy the equation. Let me check if $ x = 0 $ is a solution.

Plugging $ x = 0 $ into the left side:

$$
\frac{1}{0 - 1} + \frac{2}{0 - 2} + \frac{3}{0 - 3} + \frac{4}{0 - 4} = (-1) + (-1) + (-1) + (-1) = -4.
$$

The right side is $ 2010*0 - 4 = -4 $. So yes, $ x = 0 $ is a solution. Therefore, $ x = 0 $ is a solution. However, if there are other solutions, they may not come in pairs.

But how does this help me? Well, if I can find the sum of all solutions, maybe I can use Vieta's formula on the polynomial equation. However, as mentioned before, the equation is not a polynomial equation unless we clear denominators. Let me try that approach.

Let me denote the left side as $ L(x) $ and the right side as $ R(x) $. Then, the equation is $ L(x) = R(x) $. Let me write:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} - 2010x + 4 = 0.
$$

To eliminate denominators, multiply both sides by $ (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. Let me denote $ D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. Then:

$$
D(x) \left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) - D(x)(2010x - 4) = 0.
$$

Simplifying the first term:

Each term in the sum will have $ D(x) $ multiplied by the respective fraction. For example, the first term is $ D(x)/(x^2 - 1) $, which is $ (x^2 - 2)(x^2 - 3)(x^2 - 4) $. Similarly for the others. Therefore, the entire left side becomes:

$$
(x^2 - 2)(x^2 - 3)(x^2 - 4) + 2(x^2 - 1)(x^2 - 3)(x^2 - 4) + 3(x^2 - 1)(x^2 - 2)(x^2 - 4) + 4(x^2 - 1)(x^2 - 2)(x^2 - 3) - (2010x - 4)D(x) = 0.
$$

This seems very complicated. However, perhaps there's a way to simplify this expression. Let me denote $ y = x^2 $, so that $ D(x) = (y - 1)(y - 2)(y - 3)(y - 4) $. Then, the first part of the equation becomes:

$$
(y - 2)(y - 3)(y - 4) + 2(y - 1)(y - 3)(y - 4) + 3(y - 1)(y - 2)(y - 4) + 4(y - 1)(y - 2)(y - 3).
$$

Let me compute each term:

First term: $ (y - 2)(y - 3)(y - 4) $

Second term: $ 2(y - 1)(y - 3)(y - 4) $

Third term: $ 3(y - 1)(y - 2)(y - 4) $

Fourth term: $ 4(y - 1)(y - 2)(y - 3) $

Let me compute each of these terms. Maybe there's a pattern here. Let me denote $ A = (y - 1)(y - 2)(y - 3)(y - 4) $. Wait, but the first four terms are parts of A. Let me think.

Alternatively, note that the first four terms are similar to expanding $ (y - 1)(y - 2)(y - 3)(y - 4) $, but with coefficients. Let me think. Let me consider that if I have:

Let me denote $ S = (y - 1)(y - 2)(y - 3)(y - 4) $. Then, if I expand S, it's a quartic polynomial. However, the first four terms in the equation above are:

Term1: (y - 2)(y - 3)(y - 4) = S / (y - 1)

Term2: 2(y - 1)(y - 3)(y - 4) = 2 * S / (y - 2)

Term3: 3(y - 1)(y - 2)(y - 4) = 3 * S / (y - 3)

Term4: 4(y - 1)(y - 2)(y - 3) = 4 * S / (y - 4)

Therefore, the entire first part is:

S/(y - 1) + 2S/(y - 2) + 3S/(y - 3) + 4S/(y - 4)

But S is (y - 1)(y - 2)(y - 3)(y - 4). Therefore, the first part is:

S [ 1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4) ].

Wait, but that seems circular. Wait, no. Wait, the original left side before multiplying by D(x) was:

[1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4)].

But when we multiplied by D(x), we get S [ 1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4) ].

But that's equal to the original left side multiplied by D(x). However, this seems like going in circles. Maybe there's another way.

Alternatively, perhaps I can consider that the first four terms (before subtracting the other part) can be combined. Let me try expanding them.

Let me compute each term:

First term: (y - 2)(y - 3)(y - 4)

Let me compute (y - 2)(y - 3) = y² - 5y + 6. Then multiply by (y - 4):

(y² - 5y + 6)(y - 4) = y³ - 4y² - 5y² + 20y + 6y - 24 = y³ - 9y² + 26y - 24.

Second term: 2(y - 1)(y - 3)(y - 4)

First compute (y - 1)(y - 3) = y² - 4y + 3. Multiply by (y - 4):

(y² - 4y + 3)(y - 4) = y³ - 4y² - 4y² + 16y + 3y - 12 = y³ - 8y² + 19y - 12. Multiply by 2: 2y³ - 16y² + 38y - 24.

Third term: 3(y - 1)(y - 2)(y - 4)

Compute (y - 1)(y - 2) = y² - 3y + 2. Multiply by (y - 4):

(y² - 3y + 2)(y - 4) = y³ - 4y² - 3y² + 12y + 2y - 8 = y³ - 7y² + 14y - 8. Multiply by 3: 3y³ - 21y² + 42y - 24.

Fourth term: 4(y - 1)(y - 2)(y - 3)

Compute (y - 1)(y - 2) = y² - 3y + 2. Multiply by (y - 3):

(y² - 3y + 2)(y - 3) = y³ - 3y² - 3y² + 9y + 2y - 6 = y³ - 6y² + 11y - 6. Multiply by 4: 4y³ - 24y² + 44y - 24.

Now, adding all four terms together:

First term: y³ - 9y² + 26y - 24

Second term: + 2y³ - 16y² + 38y - 24

Third term: + 3y³ - 21y² + 42y - 24

Fourth term: + 4y³ - 24y² + 44y - 24

Adding coefficients:

For y³: 1 + 2 + 3 + 4 = 10

For y²: -9 -16 -21 -24 = -70

For y: 26 + 38 + 42 + 44 = 150

Constants: -24 -24 -24 -24 = -96

So the first part simplifies to:

10y³ - 70y² + 150y - 96.

Now, the entire equation after multiplying by D(x) is:

[10y³ - 70y² + 150y - 96] - (2010x - 4) * D(x) = 0.

But D(x) is (x² - 1)(x² - 2)(x² - 3)(x² - 4). However, since y = x², D(x) = (y - 1)(y - 2)(y - 3)(y - 4). Let me denote that as S(y) = (y - 1)(y - 2)(y - 3)(y - 4). Therefore, the equation becomes:

10y³ - 70y² + 150y - 96 - (2010x - 4) * S(y) = 0.

But this still has both x and y. Since y = x², we can write x in terms of y, but that might not be helpful. However, note that S(y) is a quartic in y, so multiplying by (2010x - 4) would give a term that is a quartic in y multiplied by x, which complicates things.

Alternatively, maybe I can express x in terms of y? But x = ±√y, which would introduce square roots. This seems messy.

Alternatively, perhaps I can consider that the entire equation is in terms of x, so maybe I need to express everything in terms of x. Let me try that.

But S(y) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the equation is:

10x^6 - 70x^4 + 150x² - 96 - (2010x - 4)(x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.

Wait, because y = x², so y³ = x^6, y² = x^4, y = x². Therefore, the first part is 10x^6 - 70x^4 + 150x² - 96.

So the entire equation is:

10x^6 - 70x^4 + 150x² - 96 - (2010x - 4)(x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.

This is a polynomial equation in x. Let me denote this as P(x) = 0, where P(x) is the above expression.

Now, the problem is that this polynomial is of degree 9, since (x² - 1)...(x² - 4) is degree 8, multiplied by 2010x gives degree 9, and the other term is degree 6. Therefore, the leading term is -2010x*(x²)^4 = -2010x^9. Therefore, the polynomial is degree 9.

But Vieta's formula tells us that the sum of all roots of P(x) = 0 is equal to - (coefficient of x^8) / (coefficient of x^9). However, we need to compute the coefficient of x^8 and x^9.

But this seems very tedious. However, maybe there's a smarter way. Let me think.

Alternatively, note that the original equation had x = 0 as a solution. Let me check if x = 0 is a root of P(x).

Let me compute P(0):

First part: 10*(0)^6 - 70*(0)^4 + 150*(0)^2 - 96 = -96.

Second part: -(2010*0 - 4)*(0 - 1)(0 - 2)(0 - 3)(0 - 4) = -(-4)*(-1)(-2)(-3)(-4). Let me compute that:

First, compute (-4) * (-1)(-2)(-3)(-4). Let me compute step by step:

(-1)(-2) = 2; 2*(-3) = -6; -6*(-4) = 24. Then, (-4)*24 = -96. Therefore, the second part is -(-96) = 96.

Therefore, P(0) = -96 + 96 = 0. So x = 0 is indeed a root.

But how does this help? Well, if I can find the coefficient of x^9 and x^8 in P(x), then I can compute the sum of all roots. However, computing these coefficients manually would be very time-consuming. However, maybe there's a way to find them without expanding the entire polynomial.

Let me consider that P(x) = [10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x), where S(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4).

Therefore, the coefficient of x^9 in P(x) is the coefficient of x^9 in - (2010x - 4) * S(x). Similarly, the coefficient of x^8 is the coefficient of x^8 in - (2010x - 4) * S(x) plus the coefficient of x^8 in the first part.

But the first part, 10x^6 - 70x^4 + 150x² - 96, has no x^8 term, so the coefficient of x^8 in P(x) is just the coefficient of x^8 in - (2010x - 4) * S(x).

Similarly, the coefficient of x^9 in P(x) is the coefficient of x^9 in - (2010x - 4) * S(x).

Therefore, let me focus on computing the coefficients of x^9 and x^8 in - (2010x - 4) * S(x).

First, let me find S(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me denote S(x) as a polynomial. Let me compute S(x):

First, compute (x² - 1)(x² - 2) = x^4 - 3x² + 2.

Then, compute (x² - 3)(x² - 4) = x^4 - 7x² + 12.

Now, multiply these two quartic polynomials:

(x^4 - 3x² + 2)(x^4 - 7x² + 12).

Let me compute this:

Multiply term by term:

First, x^4 * x^4 = x^8.

x^4 * (-7x²) = -7x^6.

x^4 * 12 = 12x^4.

Then, -3x² * x^4 = -3x^6.

-3x² * (-7x²) = 21x^4.

-3x² * 12 = -36x².

Then, 2 * x^4 = 2x^4.

2 * (-7x²) = -14x².

2 * 12 = 24.

Now, combine like terms:

x^8 + (-7x^6 - 3x^6) + (12x^4 + 21x^4 + 2x^4) + (-36x² -14x²) + 24.

Calculating each:

x^8 term: 1x^8.

x^6 term: -10x^6.

x^4 term: 35x^4.

x² term: -50x².

Constant term: 24.

Therefore, S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Therefore, S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24.

Now, let me compute - (2010x - 4) * S(x). Let me denote this as -2010x * S(x) + 4 * S(x).

Therefore, the coefficient of x^9 in this expression is the coefficient of x^9 in -2010x * S(x) plus the coefficient of x^9 in 4 * S(x). However, 4 * S(x) is a degree 8 polynomial, so it has no x^9 term. Therefore, the coefficient of x^9 is just the coefficient of x^9 in -2010x * S(x).

Similarly, the coefficient of x^9 in -2010x * S(x) is -2010 times the coefficient of x^8 in S(x).

From S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24, the coefficient of x^8 is 1. Therefore, the coefficient of x^9 is -2010 * 1 = -2010.

Similarly, the coefficient of x^8 in - (2010x - 4) * S(x) is the coefficient of x^8 in -2010x * S(x) plus the coefficient of x^8 in 4 * S(x).

First, coefficient of x^8 in -2010x * S(x): this is -2010 times the coefficient of x^7 in S(x). But S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24, so there is no x^7 term. Therefore, the coefficient is 0.

Then, the coefficient of x^8 in 4 * S(x) is 4 times the coefficient of x^8 in S(x), which is 4 * 1 = 4. Therefore, the total coefficient of x^8 in - (2010x - 4) * S(x) is 0 + 4 = 4.

Therefore, going back to P(x):

The coefficient of x^9 in P(x) is -2010, and the coefficient of x^8 in P(x) is 4.

But remember, P(x) is equal to [10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x). Therefore, the coefficient of x^8 in P(x) is 4 (from the second part) plus the coefficient of x^8 in the first part. However, the first part is 10x^6 - 70x^4 + 150x² - 96, which has no x^8 term. Therefore, the coefficient of x^8 in P(x) is 4.

Therefore, the polynomial P(x) is a degree 9 polynomial with leading term -2010x^9 and coefficient of x^8 is 4. Therefore, by Vieta's formula, the sum of all roots of P(x) = 0 is equal to - (coefficient of x^8) / (coefficient of x^9) = - (4) / (-2010) = 4 / 2010 = 2 / 1005.

But wait, hold on. Vieta's formula says that for a polynomial of the form a_n x^n + ... + a_0 = 0, the sum of roots is -a_{n-1}/a_n.

In our case, the polynomial is P(x) = -2010x^9 + ... + 4x^8 + ... So, the coefficient of x^9 is -2010, and the coefficient of x^8 is 4. Therefore, the sum of all roots is - (coefficient of x^8) / (coefficient of x^9) = - (4) / (-2010) = 4 / 2010 = 2 / 1005.

But wait, this is the sum of all roots of P(x) = 0. However, we need to remember that P(x) was obtained by multiplying the original equation by D(x), which is (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the roots of P(x) = 0 include the roots of the original equation, but also include the roots of D(x) = 0, which are x = ±1, ±√2, ±√3, ±2. However, these roots are excluded from the original equation because they make denominators zero. Therefore, the roots of P(x) = 0 are the solutions to the original equation plus the roots of D(x) = 0.

Therefore, the sum of all roots of P(x) = 0 is equal to the sum of the solutions of the original equation plus the sum of the roots of D(x) = 0. Therefore, if I denote S_total as the sum of all roots of P(x) = 0, which is 2/1005, and S_excluded as the sum of roots of D(x) = 0, then the sum of solutions of the original equation is S_total - S_excluded.

Therefore, I need to compute S_excluded, which is the sum of roots of D(x) = 0.

But D(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me note that D(x) is a degree 8 polynomial. The roots are x = ±1, ±√2, ±√3, ±2. Therefore, the sum of roots is 0, because for each root r, -r is also a root. Therefore, they cancel out.

For example, 1 and -1, √2 and -√2, etc. Therefore, the sum of all roots of D(x) = 0 is 0.

Therefore, the sum of solutions of the original equation is S_total - S_excluded = (2/1005) - 0 = 2/1005.

But wait, hold on. However, we need to check if any of the roots of D(x) = 0 are also roots of the original equation. Because when we multiplied both sides by D(x), we might have introduced extraneous roots. However, the original equation is undefined at those roots, so they cannot be solutions. Therefore, they are not included in the original equation's solutions. Therefore, the sum of solutions of the original equation is indeed S_total - S_excluded = 2/1005.

But let me check if x = 0 is included in the roots of P(x). Earlier, we saw that x = 0 is a root of P(x). However, x = 0 is not a root of D(x) = 0, since D(0) = (0 - 1)(0 - 2)(0 - 3)(0 - 4) = (-1)(-2)(-3)(-4) = 24 ≠ 0. Therefore, x = 0 is a valid solution.

But since S_excluded is the sum of roots of D(x) = 0, which is 0, and S_total is 2/1005, then the sum of solutions of the original equation is 2/1005.

But let me check if there are any other roots of D(x) = 0 that might have been introduced. However, since D(x) = 0 corresponds to x^2 = 1, 2, 3, 4, which are excluded from the original equation, so they are not solutions. Therefore, the sum of solutions of the original equation is indeed 2/1005.

But let me check if there's a mistake in the calculation of the coefficients.

Earlier, I found that the coefficient of x^9 in P(x) is -2010, and the coefficient of x^8 is 4. Therefore, the sum of roots is - (4)/(-2010) = 4/2010 = 2/1005.

But let me verify the calculation of the coefficients again.

First, S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Then, - (2010x - 4) * S(x) = -2010x * S(x) + 4 * S(x).

The coefficient of x^9 in -2010x * S(x) is -2010 times the coefficient of x^8 in S(x). Since S(x) has coefficient 1 for x^8, that term is -2010 * 1 = -2010.

The coefficient of x^8 in 4 * S(x) is 4 * 1 = 4.

Therefore, the coefficient of x^8 in - (2010x - 4) * S(x) is 4.

Then, the first part of P(x) is 10x^6 - 70x^4 + 150x² - 96, which has no x^8 term. Therefore, the coefficient of x^8 in P(x) is 4.

Therefore, the sum of roots is indeed 4 / 2010 = 2 / 1005.

But let me check if there's a mistake in the sign. Vieta's formula says that for a polynomial a_n x^n + ... + a_0 = 0, the sum of roots is -a_{n-1}/a_n.

In our case, the polynomial P(x) is:

-2010x^9 + ... + 4x^8 + ...

Therefore, a_n = -2010 (coefficient of x^9), and a_{n-1} = 4 (coefficient of x^8). Therefore, the sum of roots is -a_{n-1}/a_n = -4 / (-2010) = 4 / 2010 = 2 / 1005.

Yes, that seems correct.

But let me check if there's a mistake in the calculation of S(x). Let me recompute S(x):

Earlier, I had:

(x^4 - 3x² + 2)(x^4 - 7x² + 12) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Let me verify this multiplication again.

Multiply (x^4 - 3x² + 2) by (x^4 - 7x² + 12):

First, x^4 * x^4 = x^8.

x^4 * (-7x²) = -7x^6.

x^4 * 12 = 12x^4.

Then, -3x² * x^4 = -3x^6.

-3x² * (-7x²) = 21x^4.

-3x² * 12 = -36x².

Then, 2 * x^4 = 2x^4.

2 * (-7x²) = -14x².

2 * 12 = 24.

Now, combining terms:

x^8 + (-7x^6 - 3x^6) + (12x^4 + 21x^4 + 2x^4) + (-36x² -14x²) + 24.

Calculates to:

x^8 - 10x^6 + 35x^4 - 50x² + 24.

Yes, that's correct.

Therefore, the calculation of S(x) is correct.

Therefore, the coefficients of P(x) are correct.

Therefore, the sum of all roots of P(x) is 2/1005, and since the excluded roots sum to zero, the sum of solutions of the original equation is 2/1005.

But let me check if there's a possibility that x = 0 is a multiple root or something. However, since we already confirmed that x = 0 is a root, and the other roots are from the excluded ones, which don't affect the sum.

Alternatively, maybe there's a mistake in assuming that all roots of P(x) except those from D(x) = 0 are solutions to the original equation. However, when we multiplied both sides by D(x), we could have introduced extraneous roots where the original equation might have had solutions that are not in the original domain. However, since the original equation is undefined at the roots of D(x) = 0, those roots are not solutions. Therefore, the other roots of P(x) = 0 are solutions to the original equation.

Therefore, the sum of all complex solutions is 2/1005.

But let me check if 2/1005 can be simplified. 1005 divided by 5 is 201, so 1005 = 5 * 201 = 5 * 3 * 67. 2 and 1005 share no common factors, so 2/1005 is in simplest terms.

But let me check if the problem says "complex solutions", so we need to include all roots, even if they are real or complex. However, since we already accounted for all roots via Vieta's formula, and the excluded roots sum to zero, the answer is 2/1005.

But let me check with another approach. Let me think about the original equation again.

Original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Let me consider that if I denote $ f(x) = \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} $, then the equation is $ f(x) = 2010x - 4 $.

Note that $ f(x) $ is an even function, as mentioned before. However, the right side is not even. However, if we consider that for each solution $ x $, there might be a corresponding solution $ -x $, but only if $ f(x) = 2010x - 4 $ and $ f(-x) = -2010x - 4 $. Therefore, unless $ 2010x - 4 = -2010x - 4 $, which implies $ 4020x = 0 $, i.e., $ x = 0 $, there is no symmetry. Therefore, except for $ x = 0 $, solutions don't come in pairs.

But if we suppose that the sum of all solutions is S, then if there are solutions other than 0, they might not come in pairs. However, according to our previous calculation, the sum is 2/1005. However, if we consider that the equation is not symmetric, how can we be sure?

But according to the polynomial approach, the sum is 2/1005. However, let me check with x = 0. Since x = 0 is a solution, and the rest of the solutions, if they exist, would have their sum as 2/1005 - 0 = 2/1005.

Alternatively, maybe there's a mistake in the sign. Let me check Vieta again.

The polynomial P(x) is equal to:

[10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x) = 0.

Which is:

-2010x * S(x) + 4 * S(x) + 10x^6 - 70x^4 + 150x² - 96 = 0.

Therefore, the leading term is -2010x * x^8 = -2010x^9.

Therefore, the polynomial is of degree 9, with leading coefficient -2010.

The coefficient of x^8 is 4, as calculated.

Therefore, the sum of roots is - (coefficient of x^8) / (coefficient of x^9) = -4 / (-2010) = 4 / 2010 = 2 / 1005.

Yes, that's correct.

Therefore, the answer should be 2/1005.

But let me check if there's a possible miscalculation in the coefficient of x^8.

Earlier, I said that the coefficient of x^8 in - (2010x - 4) * S(x) is 4. Let me verify this again.

The term - (2010x - 4) * S(x) is equal to -2010x * S(x) + 4 * S(x).

The coefficient of x^8 in -2010x * S(x) is -2010 times the coefficient of x^7 in S(x). But S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24. There is no x^7 term, so the coefficient is 0.

The coefficient of x^8 in 4 * S(x) is 4 times the coefficient of x^8 in S(x), which is 4 * 1 = 4.

Therefore, total coefficient of x^8 is 0 + 4 = 4.

Therefore, correct.

Therefore, the sum of all roots of P(x) is 2/1005.

Since the excluded roots sum to zero, the sum of solutions of the original equation is 2/1005.

Therefore, the answer is 2/1005.

But let me check if 2/1005 reduces. 1005 divided by 5 is 201, 201 divided by 3 is 67. So 1005 = 5 * 3 * 67. 2 and 1005 share no common factors, so 2/1005 is the simplified form.

Alternatively, maybe the problem expects the answer in some other form? Let me check if 2/1005 can be simplified further. 2 and 1005 are coprime? 1005 ÷ 2 = 502.5, so no. Therefore, yes, 2/1005 is simplest.

Alternatively, maybe I made a mistake in the sign? Let me check Vieta again.

The polynomial is P(x) = -2010x^9 + ... + 4x^8 + ...

So, in standard form, it's a_9 x^9 + a_8 x^8 + ... + a_0 = 0, where a_9 = -2010, a_8 = 4.

Sum of roots is -a_8 / a_9 = -4 / (-2010) = 4/2010 = 2/1005.

Yes, correct.

Therefore, I think this is the correct answer.

**Final Answer**
\boxed{\dfrac{2}{1005}}

To solve the equation

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4,
$$

we aim to find the **sum of all complex solutions**.

---

### Step 1: Recognize the Structure

The left-hand side is a sum of rational functions in $ x^2 $, and the right-hand side is a linear function in $ x $. This suggests that the equation is not a polynomial, but we can attempt to convert it into a polynomial by eliminating denominators.

Let us define:

$$
y = x^2.
$$

Then the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

However, the right-hand side still contains $ x $, not $ y $, so we cannot directly express the equation in terms of $ y $ alone. Therefore, we proceed by eliminating denominators.

---

### Step 2: Eliminate Denominators

Let us denote the common denominator as:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

Multiplying both sides of the equation by $ D(x) $, we get:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) D(x) = (2010x - 4) D(x).
$$

This results in a polynomial equation in $ x $, which we denote as $ P(x) = 0 $, where:

$$
P(x) = \left( \text{left side} \right) D(x) - (2010x - 4) D(x) = 0.
$$

This is a **degree 9 polynomial** in $ x $, since $ D(x) $ is degree 8 and $ (2010x - 4) D(x) $ is degree 9.

---

### Step 3: Use Vieta's Formula

Vieta's formula tells us that the **sum of all roots** of a polynomial $ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0 $ is:

$$
\text{Sum of roots} = -\frac{a_{n-1}}{a_n}.
$$

We now compute the leading terms of $ P(x) $ to find the coefficients of $ x^9 $ and $ x^8 $.

Let’s denote:

$$
S(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

We compute:

$$
S(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) = x^8 - 10x^6 + 35x^4 - 50x^2 + 24.
$$

Then:

$$
P(x) = \left( \text{polynomial of degree 6} \right) - (2010x - 4) S(x).
$$

The leading term of $ P(x) $ comes from $ -2010x \cdot x^8 = -2010x^9 $, so the coefficient of $ x^9 $ is $ -2010 $.

The coefficient of $ x^8 $ comes from:

- $ -2010x \cdot x^7 $: but $ S(x) $ has no $ x^7 $ term, so this is 0.
- $ 4 \cdot x^8 $: from $ 4 \cdot S(x) $, which has a leading term $ 4x^8 $.

So the coefficient of $ x^8 $ is $ 4 $.

Thus, the sum of all roots of $ P(x) = 0 $ is:

$$
\text{Sum of roots} = -\frac{4}{-2010} = \frac{4}{2010} = \frac{2}{1005}.
$$

---

### Step 4: Exclude Invalid Roots

However, the original equation is undefined when $ x^2 = 1, 2, 3, 4 $, i.e., when $ x = \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2 $. These values make the denominators zero, so they are **not** valid solutions.

Let’s denote the set of such excluded values as the roots of $ D(x) = 0 $, which is:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

The roots of $ D(x) = 0 $ are $ \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2 $, and their sum is:

$$
1 + (-1) + \sqrt{2} + (-\sqrt{2}) + \sqrt{3} + (-\sqrt{3}) + 2 + (-2) = 0.
$$

So the sum of the excluded roots is **zero**.

---

### Step 5: Final Answer

The sum of all **complex solutions** of the original equation is the sum of all roots of $ P(x) = 0 $, **excluding** the roots of $ D(x) = 0 $, which sum to zero. Therefore, the sum of the complex solutions is:

$$
\boxed{\dfrac{2}{1005}}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of all complex solutions to this equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Hmm, complex solutions... So, I need to consider all roots of this equation, even if they are complex numbers. The question is asking for their sum.

First, let me think about how to approach this equation. It has rational expressions on the left side and a linear term on the right. The denominators are all quadratic in terms of $ x^2 $, so maybe I can simplify the equation by substituting $ y = x^2 $. Let me try that.

Let me set $ y = x^2 $. Then the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

Wait, but there's still an $ x $ term on the right side. If I substitute $ y = x^2 $, then $ x $ is still present. Maybe I need another substitution or approach. Alternatively, perhaps I can consider that the left side is a function of $ y $, and the right side is a function of $ x $. But since $ y = x^2 $, maybe I can express everything in terms of $ y $, but the problem is that the right side has $ x $, not $ y $.

Alternatively, maybe I can rearrange the equation to bring all terms to one side and then find a common denominator? But that might get complicated. Let me think.

Alternatively, maybe I can consider that if I have an equation involving $ x $, and I need to find the sum of all complex solutions. For polynomial equations, the sum of roots can be found using Vieta's formula. However, this equation isn't a polynomial equation because of the denominators. But perhaps if I can manipulate it into a polynomial equation, then Vieta's formula can be applied.

So, the plan is: first, eliminate denominators by multiplying both sides by the common denominator, which would be $ (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. However, this would result in a high-degree polynomial equation. Let me check the degrees.

Each denominator is quadratic, so multiplying them would give a degree 8 polynomial. However, the right side is $ 2010x - 4 $, so when multiplied by the common denominator, the right side would be $ (2010x - 4)(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. That would be a degree 9 polynomial. Therefore, the entire equation would be a degree 9 polynomial equation. However, the left side, when multiplied by the common denominator, would be a polynomial of degree 8. Therefore, moving everything to one side would result in a degree 9 polynomial.

But solving a degree 9 polynomial is not straightforward. However, the question is not asking for the roots, but the sum of all complex solutions. Maybe there's a trick here.

Wait, Vieta's formula tells us that the sum of the roots of a polynomial equation $ a_n x^n + \dots + a_0 = 0 $ is $ -a_{n-1}/a_n $. So, if I can write the equation as a polynomial equation, then I can find the sum of the roots. However, the problem is that the original equation has denominators, so when we multiply through by the common denominator, we might introduce extraneous roots (the ones that make denominators zero), but since we are considering complex solutions, maybe we need to be careful. However, the original equation is undefined when $ x^2 = 1, 2, 3, 4 $, so those values of $ x $ are excluded. Therefore, when we multiply both sides by the common denominator, we have to ensure that we don't include those roots. However, since we are looking for the sum of all complex solutions, which are the roots of the polynomial equation excluding those that make denominators zero. But maybe the sum of all roots of the polynomial equation (including those excluded) minus the sum of the excluded roots would give the desired sum. However, this seems complicated.

Alternatively, maybe there's a symmetry or substitution that can be used here. Let me think again.

Looking back at the original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Note that the left side is a function of $ x^2 $, while the right side is a linear function of $ x $. If I consider that the left side is even in $ x $, because replacing $ x $ with $ -x $ doesn't change $ x^2 $, so the left side is even. However, the right side is $ 2010x - 4 $, which is an odd function plus a constant. Therefore, the entire equation is not symmetric. However, if we consider that for each solution $ x $, there might be a corresponding solution $ -x $, but only if the right side satisfies some condition. Let me check.

Suppose $ x $ is a solution. Then, substituting $ -x $ into the equation:

Left side becomes the same as for $ x $, since $ (-x)^2 = x^2 $. The right side becomes $ 2010(-x) - 4 = -2010x - 4 $. Therefore, unless $ 2010x - 4 = -2010x - 4 $, which would imply $ 4020x = 0 $, i.e., $ x = 0 $, the equation is not symmetric. Therefore, except for $ x = 0 $, solutions don't come in pairs $ x $ and $ -x $. However, if $ x = 0 $ is a solution, then it would satisfy the equation. Let me check if $ x = 0 $ is a solution.

Plugging $ x = 0 $ into the left side:

$$
\frac{1}{0 - 1} + \frac{2}{0 - 2} + \frac{3}{0 - 3} + \frac{4}{0 - 4} = (-1) + (-1) + (-1) + (-1) = -4.
$$

The right side is $ 2010*0 - 4 = -4 $. So yes, $ x = 0 $ is a solution. Therefore, $ x = 0 $ is a solution. However, if there are other solutions, they may not come in pairs.

But how does this help me? Well, if I can find the sum of all solutions, maybe I can use Vieta's formula on the polynomial equation. However, as mentioned before, the equation is not a polynomial equation unless we clear denominators. Let me try that approach.

Let me denote the left side as $ L(x) $ and the right side as $ R(x) $. Then, the equation is $ L(x) = R(x) $. Let me write:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} - 2010x + 4 = 0.
$$

To eliminate denominators, multiply both sides by $ (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. Let me denote $ D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) $. Then:

$$
D(x) \left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) - D(x)(2010x - 4) = 0.
$$

Simplifying the first term:

Each term in the sum will have $ D(x) $ multiplied by the respective fraction. For example, the first term is $ D(x)/(x^2 - 1) $, which is $ (x^2 - 2)(x^2 - 3)(x^2 - 4) $. Similarly for the others. Therefore, the entire left side becomes:

$$
(x^2 - 2)(x^2 - 3)(x^2 - 4) + 2(x^2 - 1)(x^2 - 3)(x^2 - 4) + 3(x^2 - 1)(x^2 - 2)(x^2 - 4) + 4(x^2 - 1)(x^2 - 2)(x^2 - 3) - (2010x - 4)D(x) = 0.
$$

This seems very complicated. However, perhaps there's a way to simplify this expression. Let me denote $ y = x^2 $, so that $ D(x) = (y - 1)(y - 2)(y - 3)(y - 4) $. Then, the first part of the equation becomes:

$$
(y - 2)(y - 3)(y - 4) + 2(y - 1)(y - 3)(y - 4) + 3(y - 1)(y - 2)(y - 4) + 4(y - 1)(y - 2)(y - 3).
$$

Let me compute each term:

First term: $ (y - 2)(y - 3)(y - 4) $

Second term: $ 2(y - 1)(y - 3)(y - 4) $

Third term: $ 3(y - 1)(y - 2)(y - 4) $

Fourth term: $ 4(y - 1)(y - 2)(y - 3) $

Let me compute each of these terms. Maybe there's a pattern here. Let me denote $ A = (y - 1)(y - 2)(y - 3)(y - 4) $. Wait, but the first four terms are parts of A. Let me think.

Alternatively, note that the first four terms are similar to expanding $ (y - 1)(y - 2)(y - 3)(y - 4) $, but with coefficients. Let me think. Let me consider that if I have:

Let me denote $ S = (y - 1)(y - 2)(y - 3)(y - 4) $. Then, if I expand S, it's a quartic polynomial. However, the first four terms in the equation above are:

Term1: (y - 2)(y - 3)(y - 4) = S / (y - 1)

Term2: 2(y - 1)(y - 3)(y - 4) = 2 * S / (y - 2)

Term3: 3(y - 1)(y - 2)(y - 4) = 3 * S / (y - 3)

Term4: 4(y - 1)(y - 2)(y - 3) = 4 * S / (y - 4)

Therefore, the entire first part is:

S/(y - 1) + 2S/(y - 2) + 3S/(y - 3) + 4S/(y - 4)

But S is (y - 1)(y - 2)(y - 3)(y - 4). Therefore, the first part is:

S [ 1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4) ].

Wait, but that seems circular. Wait, no. Wait, the original left side before multiplying by D(x) was:

[1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4)].

But when we multiplied by D(x), we get S [ 1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4) ].

But that's equal to the original left side multiplied by D(x). However, this seems like going in circles. Maybe there's another way.

Alternatively, perhaps I can consider that the first four terms (before subtracting the other part) can be combined. Let me try expanding them.

Let me compute each term:

First term: (y - 2)(y - 3)(y - 4)

Let me compute (y - 2)(y - 3) = y² - 5y + 6. Then multiply by (y - 4):

(y² - 5y + 6)(y - 4) = y³ - 4y² - 5y² + 20y + 6y - 24 = y³ - 9y² + 26y - 24.

Second term: 2(y - 1)(y - 3)(y - 4)

First compute (y - 1)(y - 3) = y² - 4y + 3. Multiply by (y - 4):

(y² - 4y + 3)(y - 4) = y³ - 4y² - 4y² + 16y + 3y - 12 = y³ - 8y² + 19y - 12. Multiply by 2: 2y³ - 16y² + 38y - 24.

Third term: 3(y - 1)(y - 2)(y - 4)

Compute (y - 1)(y - 2) = y² - 3y + 2. Multiply by (y - 4):

(y² - 3y + 2)(y - 4) = y³ - 4y² - 3y² + 12y + 2y - 8 = y³ - 7y² + 14y - 8. Multiply by 3: 3y³ - 21y² + 42y - 24.

Fourth term: 4(y - 1)(y - 2)(y - 3)

Compute (y - 1)(y - 2) = y² - 3y + 2. Multiply by (y - 3):

(y² - 3y + 2)(y - 3) = y³ - 3y² - 3y² + 9y + 2y - 6 = y³ - 6y² + 11y - 6. Multiply by 4: 4y³ - 24y² + 44y - 24.

Now, adding all four terms together:

First term: y³ - 9y² + 26y - 24

Second term: + 2y³ - 16y² + 38y - 24

Third term: + 3y³ - 21y² + 42y - 24

Fourth term: + 4y³ - 24y² + 44y - 24

Adding coefficients:

For y³: 1 + 2 + 3 + 4 = 10

For y²: -9 -16 -21 -24 = -70

For y: 26 + 38 + 42 + 44 = 150

Constants: -24 -24 -24 -24 = -96

So the first part simplifies to:

10y³ - 70y² + 150y - 96.

Now, the entire equation after multiplying by D(x) is:

[10y³ - 70y² + 150y - 96] - (2010x - 4) * D(x) = 0.

But D(x) is (x² - 1)(x² - 2)(x² - 3)(x² - 4). However, since y = x², D(x) = (y - 1)(y - 2)(y - 3)(y - 4). Let me denote that as S(y) = (y - 1)(y - 2)(y - 3)(y - 4). Therefore, the equation becomes:

10y³ - 70y² + 150y - 96 - (2010x - 4) * S(y) = 0.

But this still has both x and y. Since y = x², we can write x in terms of y, but that might not be helpful. However, note that S(y) is a quartic in y, so multiplying by (2010x - 4) would give a term that is a quartic in y multiplied by x, which complicates things.

Alternatively, maybe I can express x in terms of y? But x = ±√y, which would introduce square roots. This seems messy.

Alternatively, perhaps I can consider that the entire equation is in terms of x, so maybe I need to express everything in terms of x. Let me try that.

But S(y) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the equation is:

10x^6 - 70x^4 + 150x² - 96 - (2010x - 4)(x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.

Wait, because y = x², so y³ = x^6, y² = x^4, y = x². Therefore, the first part is 10x^6 - 70x^4 + 150x² - 96.

So the entire equation is:

10x^6 - 70x^4 + 150x² - 96 - (2010x - 4)(x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.

This is a polynomial equation in x. Let me denote this as P(x) = 0, where P(x) is the above expression.

Now, the problem is that this polynomial is of degree 9, since (x² - 1)...(x² - 4) is degree 8, multiplied by 2010x gives degree 9, and the other term is degree 6. Therefore, the leading term is -2010x*(x²)^4 = -2010x^9. Therefore, the polynomial is degree 9.

But Vieta's formula tells us that the sum of all roots of P(x) = 0 is equal to - (coefficient of x^8) / (coefficient of x^9). However, we need to compute the coefficient of x^8 and x^9.

But this seems very tedious. However, maybe there's a smarter way. Let me think.

Alternatively, note that the original equation had x = 0 as a solution. Let me check if x = 0 is a root of P(x).

Let me compute P(0):

First part: 10*(0)^6 - 70*(0)^4 + 150*(0)^2 - 96 = -96.

Second part: -(2010*0 - 4)*(0 - 1)(0 - 2)(0 - 3)(0 - 4) = -(-4)*(-1)(-2)(-3)(-4). Let me compute that:

First, compute (-4) * (-1)(-2)(-3)(-4). Let me compute step by step:

(-1)(-2) = 2; 2*(-3) = -6; -6*(-4) = 24. Then, (-4)*24 = -96. Therefore, the second part is -(-96) = 96.

Therefore, P(0) = -96 + 96 = 0. So x = 0 is indeed a root.

But how does this help? Well, if I can find the coefficient of x^9 and x^8 in P(x), then I can compute the sum of all roots. However, computing these coefficients manually would be very time-consuming. However, maybe there's a way to find them without expanding the entire polynomial.

Let me consider that P(x) = [10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x), where S(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4).

Therefore, the coefficient of x^9 in P(x) is the coefficient of x^9 in - (2010x - 4) * S(x). Similarly, the coefficient of x^8 is the coefficient of x^8 in - (2010x - 4) * S(x) plus the coefficient of x^8 in the first part.

But the first part, 10x^6 - 70x^4 + 150x² - 96, has no x^8 term, so the coefficient of x^8 in P(x) is just the coefficient of x^8 in - (2010x - 4) * S(x).

Similarly, the coefficient of x^9 in P(x) is the coefficient of x^9 in - (2010x - 4) * S(x).

Therefore, let me focus on computing the coefficients of x^9 and x^8 in - (2010x - 4) * S(x).

First, let me find S(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me denote S(x) as a polynomial. Let me compute S(x):

First, compute (x² - 1)(x² - 2) = x^4 - 3x² + 2.

Then, compute (x² - 3)(x² - 4) = x^4 - 7x² + 12.

Now, multiply these two quartic polynomials:

(x^4 - 3x² + 2)(x^4 - 7x² + 12).

Let me compute this:

Multiply term by term:

First, x^4 * x^4 = x^8.

x^4 * (-7x²) = -7x^6.

x^4 * 12 = 12x^4.

Then, -3x² * x^4 = -3x^6.

-3x² * (-7x²) = 21x^4.

-3x² * 12 = -36x².

Then, 2 * x^4 = 2x^4.

2 * (-7x²) = -14x².

2 * 12 = 24.

Now, combine like terms:

x^8 + (-7x^6 - 3x^6) + (12x^4 + 21x^4 + 2x^4) + (-36x² -14x²) + 24.

Calculating each:

x^8 term: 1x^8.

x^6 term: -10x^6.

x^4 term: 35x^4.

x² term: -50x².

Constant term: 24.

Therefore, S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Therefore, S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24.

Now, let me compute - (2010x - 4) * S(x). Let me denote this as -2010x * S(x) + 4 * S(x).

Therefore, the coefficient of x^9 in this expression is the coefficient of x^9 in -2010x * S(x) plus the coefficient of x^9 in 4 * S(x). However, 4 * S(x) is a degree 8 polynomial, so it has no x^9 term. Therefore, the coefficient of x^9 is just the coefficient of x^9 in -2010x * S(x).

Similarly, the coefficient of x^9 in -2010x * S(x) is -2010 times the coefficient of x^8 in S(x).

From S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24, the coefficient of x^8 is 1. Therefore, the coefficient of x^9 is -2010 * 1 = -2010.

Similarly, the coefficient of x^8 in - (2010x - 4) * S(x) is the coefficient of x^8 in -2010x * S(x) plus the coefficient of x^8 in 4 * S(x).

First, coefficient of x^8 in -2010x * S(x): this is -2010 times the coefficient of x^7 in S(x). But S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24, so there is no x^7 term. Therefore, the coefficient is 0.

Then, the coefficient of x^8 in 4 * S(x) is 4 times the coefficient of x^8 in S(x), which is 4 * 1 = 4. Therefore, the total coefficient of x^8 in - (2010x - 4) * S(x) is 0 + 4 = 4.

Therefore, going back to P(x):

The coefficient of x^9 in P(x) is -2010, and the coefficient of x^8 in P(x) is 4.

But remember, P(x) is equal to [10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x). Therefore, the coefficient of x^8 in P(x) is 4 (from the second part) plus the coefficient of x^8 in the first part. However, the first part is 10x^6 - 70x^4 + 150x² - 96, which has no x^8 term. Therefore, the coefficient of x^8 in P(x) is 4.

Therefore, the polynomial P(x) is a degree 9 polynomial with leading term -2010x^9 and coefficient of x^8 is 4. Therefore, by Vieta's formula, the sum of all roots of P(x) = 0 is equal to - (coefficient of x^8) / (coefficient of x^9) = - (4) / (-2010) = 4 / 2010 = 2 / 1005.

But wait, hold on. Vieta's formula says that for a polynomial of the form a_n x^n + ... + a_0 = 0, the sum of roots is -a_{n-1}/a_n.

In our case, the polynomial is P(x) = -2010x^9 + ... + 4x^8 + ... So, the coefficient of x^9 is -2010, and the coefficient of x^8 is 4. Therefore, the sum of all roots is - (coefficient of x^8) / (coefficient of x^9) = - (4) / (-2010) = 4 / 2010 = 2 / 1005.

But wait, this is the sum of all roots of P(x) = 0. However, we need to remember that P(x) was obtained by multiplying the original equation by D(x), which is (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the roots of P(x) = 0 include the roots of the original equation, but also include the roots of D(x) = 0, which are x = ±1, ±√2, ±√3, ±2. However, these roots are excluded from the original equation because they make denominators zero. Therefore, the roots of P(x) = 0 are the solutions to the original equation plus the roots of D(x) = 0.

Therefore, the sum of all roots of P(x) = 0 is equal to the sum of the solutions of the original equation plus the sum of the roots of D(x) = 0. Therefore, if I denote S_total as the sum of all roots of P(x) = 0, which is 2/1005, and S_excluded as the sum of roots of D(x) = 0, then the sum of solutions of the original equation is S_total - S_excluded.

Therefore, I need to compute S_excluded, which is the sum of roots of D(x) = 0.

But D(x) = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me note that D(x) is a degree 8 polynomial. The roots are x = ±1, ±√2, ±√3, ±2. Therefore, the sum of roots is 0, because for each root r, -r is also a root. Therefore, they cancel out.

For example, 1 and -1, √2 and -√2, etc. Therefore, the sum of all roots of D(x) = 0 is 0.

Therefore, the sum of solutions of the original equation is S_total - S_excluded = (2/1005) - 0 = 2/1005.

But wait, hold on. However, we need to check if any of the roots of D(x) = 0 are also roots of the original equation. Because when we multiplied both sides by D(x), we might have introduced extraneous roots. However, the original equation is undefined at those roots, so they cannot be solutions. Therefore, they are not included in the original equation's solutions. Therefore, the sum of solutions of the original equation is indeed S_total - S_excluded = 2/1005.

But let me check if x = 0 is included in the roots of P(x). Earlier, we saw that x = 0 is a root of P(x). However, x = 0 is not a root of D(x) = 0, since D(0) = (0 - 1)(0 - 2)(0 - 3)(0 - 4) = (-1)(-2)(-3)(-4) = 24 ≠ 0. Therefore, x = 0 is a valid solution.

But since S_excluded is the sum of roots of D(x) = 0, which is 0, and S_total is 2/1005, then the sum of solutions of the original equation is 2/1005.

But let me check if there are any other roots of D(x) = 0 that might have been introduced. However, since D(x) = 0 corresponds to x^2 = 1, 2, 3, 4, which are excluded from the original equation, so they are not solutions. Therefore, the sum of solutions of the original equation is indeed 2/1005.

But let me check if there's a mistake in the calculation of the coefficients.

Earlier, I found that the coefficient of x^9 in P(x) is -2010, and the coefficient of x^8 is 4. Therefore, the sum of roots is - (4)/(-2010) = 4/2010 = 2/1005.

But let me verify the calculation of the coefficients again.

First, S(x) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Then, - (2010x - 4) * S(x) = -2010x * S(x) + 4 * S(x).

The coefficient of x^9 in -2010x * S(x) is -2010 times the coefficient of x^8 in S(x). Since S(x) has coefficient 1 for x^8, that term is -2010 * 1 = -2010.

The coefficient of x^8 in 4 * S(x) is 4 * 1 = 4.

Therefore, the coefficient of x^8 in - (2010x - 4) * S(x) is 4.

Then, the first part of P(x) is 10x^6 - 70x^4 + 150x² - 96, which has no x^8 term. Therefore, the coefficient of x^8 in P(x) is 4.

Therefore, the sum of roots is indeed 4 / 2010 = 2 / 1005.

But let me check if there's a mistake in the sign. Vieta's formula says that for a polynomial a_n x^n + ... + a_0 = 0, the sum of roots is -a_{n-1}/a_n.

In our case, the polynomial P(x) is:

-2010x^9 + ... + 4x^8 + ...

Therefore, a_n = -2010 (coefficient of x^9), and a_{n-1} = 4 (coefficient of x^8). Therefore, the sum of roots is -a_{n-1}/a_n = -4 / (-2010) = 4 / 2010 = 2 / 1005.

Yes, that seems correct.

But let me check if there's a mistake in the calculation of S(x). Let me recompute S(x):

Earlier, I had:

(x^4 - 3x² + 2)(x^4 - 7x² + 12) = x^8 - 10x^6 + 35x^4 - 50x² + 24.

Let me verify this multiplication again.

Multiply (x^4 - 3x² + 2) by (x^4 - 7x² + 12):

First, x^4 * x^4 = x^8.

x^4 * (-7x²) = -7x^6.

x^4 * 12 = 12x^4.

Then, -3x² * x^4 = -3x^6.

-3x² * (-7x²) = 21x^4.

-3x² * 12 = -36x².

Then, 2 * x^4 = 2x^4.

2 * (-7x²) = -14x².

2 * 12 = 24.

Now, combining terms:

x^8 + (-7x^6 - 3x^6) + (12x^4 + 21x^4 + 2x^4) + (-36x² -14x²) + 24.

Calculates to:

x^8 - 10x^6 + 35x^4 - 50x² + 24.

Yes, that's correct.

Therefore, the calculation of S(x) is correct.

Therefore, the coefficients of P(x) are correct.

Therefore, the sum of all roots of P(x) is 2/1005, and since the excluded roots sum to zero, the sum of solutions of the original equation is 2/1005.

But let me check if there's a possibility that x = 0 is a multiple root or something. However, since we already confirmed that x = 0 is a root, and the other roots are from the excluded ones, which don't affect the sum.

Alternatively, maybe there's a mistake in assuming that all roots of P(x) except those from D(x) = 0 are solutions to the original equation. However, when we multiplied both sides by D(x), we could have introduced extraneous roots where the original equation might have had solutions that are not in the original domain. However, since the original equation is undefined at the roots of D(x) = 0, those roots are not solutions. Therefore, the other roots of P(x) = 0 are solutions to the original equation.

Therefore, the sum of all complex solutions is 2/1005.

But let me check if 2/1005 can be simplified. 1005 divided by 5 is 201, so 1005 = 5 * 201 = 5 * 3 * 67. 2 and 1005 share no common factors, so 2/1005 is in simplest terms.

But let me check if the problem says "complex solutions", so we need to include all roots, even if they are real or complex. However, since we already accounted for all roots via Vieta's formula, and the excluded roots sum to zero, the answer is 2/1005.

But let me check with another approach. Let me think about the original equation again.

Original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Let me consider that if I denote $ f(x) = \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} $, then the equation is $ f(x) = 2010x - 4 $.

Note that $ f(x) $ is an even function, as mentioned before. However, the right side is not even. However, if we consider that for each solution $ x $, there might be a corresponding solution $ -x $, but only if $ f(x) = 2010x - 4 $ and $ f(-x) = -2010x - 4 $. Therefore, unless $ 2010x - 4 = -2010x - 4 $, which implies $ 4020x = 0 $, i.e., $ x = 0 $, there is no symmetry. Therefore, except for $ x = 0 $, solutions don't come in pairs.

But if we suppose that the sum of all solutions is S, then if there are solutions other than 0, they might not come in pairs. However, according to our previous calculation, the sum is 2/1005. However, if we consider that the equation is not symmetric, how can we be sure?

But according to the polynomial approach, the sum is 2/1005. However, let me check with x = 0. Since x = 0 is a solution, and the rest of the solutions, if they exist, would have their sum as 2/1005 - 0 = 2/1005.

Alternatively, maybe there's a mistake in the sign. Let me check Vieta again.

The polynomial P(x) is equal to:

[10x^6 - 70x^4 + 150x² - 96] - (2010x - 4) * S(x) = 0.

Which is:

-2010x * S(x) + 4 * S(x) + 10x^6 - 70x^4 + 150x² - 96 = 0.

Therefore, the leading term is -2010x * x^8 = -2010x^9.

Therefore, the polynomial is of degree 9, with leading coefficient -2010.

The coefficient of x^8 is 4, as calculated.

Therefore, the sum of roots is - (coefficient of x^8) / (coefficient of x^9) = -4 / (-2010) = 4 / 2010 = 2 / 1005.

Yes, that's correct.

Therefore, the answer should be 2/1005.

But let me check if there's a possible miscalculation in the coefficient of x^8.

Earlier, I said that the coefficient of x^8 in - (2010x - 4) * S(x) is 4. Let me verify this again.

The term - (2010x - 4) * S(x) is equal to -2010x * S(x) + 4 * S(x).

The coefficient of x^8 in -2010x * S(x) is -2010 times the coefficient of x^7 in S(x). But S(x) is x^8 - 10x^6 + 35x^4 - 50x² + 24. There is no x^7 term, so the coefficient is 0.

The coefficient of x^8 in 4 * S(x) is 4 times the coefficient of x^8 in S(x), which is 4 * 1 = 4.

Therefore, total coefficient of x^8 is 0 + 4 = 4.

Therefore, correct.

Therefore, the sum of all roots of P(x) is 2/1005.

Since the excluded roots sum to zero, the sum of solutions of the original equation is 2/1005.

Therefore, the answer is 2/1005.

But let me check if 2/1005 reduces. 1005 divided by 5 is 201, 201 divided by 3 is 67. So 1005 = 5 * 3 * 67. 2 and 1005 share no common factors, so 2/1005 is the simplified form.

Alternatively, maybe the problem expects the answer in some other form? Let me check if 2/1005 can be simplified further. 2 and 1005 are coprime? 1005 ÷ 2 = 502.5, so no. Therefore, yes, 2/1005 is simplest.

Alternatively, maybe I made a mistake in the sign? Let me check Vieta again.

The polynomial is P(x) = -2010x^9 + ... + 4x^8 + ...

So, in standard form, it's a_9 x^9 + a_8 x^8 + ... + a_0 = 0, where a_9 = -2010, a_8 = 4.

Sum of roots is -a_8 / a_9 = -4 / (-2010) = 4/2010 = 2/1005.

Yes, correct.

Therefore, I think this is the correct answer.

**Final Answer**
\boxed{\dfrac{2}{1005}}

To solve the equation

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4,
$$

we aim to find the **sum of all complex solutions**.

---

### Step 1: Recognize the Structure

The left-hand side is a sum of rational functions in $ x^2 $, and the right-hand side is a linear function in $ x $. This suggests that the equation is not a polynomial, but we can attempt to convert it into a polynomial by eliminating denominators.

Let us define:

$$
y = x^2.
$$

Then the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

However, the right-hand side still contains $ x $, not $ y $, so we cannot directly express the equation in terms of $ y $ alone. Therefore, we proceed by eliminating denominators.

---

### Step 2: Eliminate Denominators

Let us denote the common denominator as:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

Multiplying both sides of the equation by $ D(x) $, we get:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) D(x) = (2010x - 4) D(x).
$$

This results in a polynomial equation in $ x $, which we denote as $ P(x) = 0 $, where:

$$
P(x) = \left( \text{left side} \right) D(x) - (2010x - 4) D(x) = 0.
$$

This is a **degree 9 polynomial** in $ x $, since $ D(x) $ is degree 8 and $ (2010x - 4) D(x) $ is degree 9.

---

### Step 3: Use Vieta's Formula

Vieta's formula tells us that the **sum of all roots** of a polynomial $ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0 $ is:

$$
\text{Sum of roots} = -\frac{a_{n-1}}{a_n}.
$$

We now compute the leading terms of $ P(x) $ to find the coefficients of $ x^9 $ and $ x^8 $.

Let’s denote:

$$
S(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

We compute:

$$
S(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) = x^8 - 10x^6 + 35x^4 - 50x^2 + 24.
$$

Then:

$$
P(x) = \left( \text{polynomial of degree 6} \right) - (2010x - 4) S(x).
$$

The leading term of $ P(x) $ comes from $ -2010x \cdot x^8 = -2010x^9 $, so the coefficient of $ x^9 $ is $ -2010 $.

The coefficient of $ x^8 $ comes from:

- $ -2010x \cdot x^7 $: but $ S(x) $ has no $ x^7 $ term, so this is 0.
- $ 4 \cdot x^8 $: from $ 4 \cdot S(x) $, which has a leading term $ 4x^8 $.

So the coefficient of $ x^8 $ is $ 4 $.

Thus, the sum of all roots of $ P(x) = 0 $ is:

$$
\text{Sum of roots} = -\frac{4}{-2010} = \frac{4}{2010} = \frac{2}{1005}.
$$

---

### Step 4: Exclude Invalid Roots

However, the original equation is undefined when $ x^2 = 1, 2, 3, 4 $, i.e., when $ x = \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2 $. These values make the denominators zero, so they are **not** valid solutions.

Let’s denote the set of such excluded values as the roots of $ D(x) = 0 $, which is:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
$$

The roots of $ D(x) = 0 $ are $ \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2 $, and their sum is:

$$
1 + (-1) + \sqrt{2} + (-\sqrt{2}) + \sqrt{3} + (-\sqrt{3}) + 2 + (-2) = 0.
$$

So the sum of the excluded roots is **zero**.

---

### Step 5: Final Answer

The sum of all **complex solutions** of the original equation is the sum of all roots of $ P(x) = 0 $, **excluding** the roots of $ D(x) = 0 $, which sum to zero. Therefore, the sum of the complex solutions is:

$$
\boxed{\dfrac{2}{1005}}
$$

Solution:
We can add 4 to both sides to get
\[\frac{1}{x^2 - 1} + 1 + \frac{2}{x^2 - 2} + 1 + \frac{3}{x^2 - 3} + 1 + \frac{4}{x^2 - 4} + 1 = 2010x.\]This simplifies to
\[\frac{x^2}{x^2 - 1} + \frac{x^2}{x^2 - 2} + \frac{x^2}{x^2 - 3} + \frac{x^2}{x^2 - 4} = 2010x.\]We see that $x = 0$ is a solution (which does not affect our sum).  Otherwise, we can divide both sides by $x$:
\[\frac{x}{x^2 - 1} + \frac{x}{x^2 - 2} + \frac{x}{x^2 - 3} + \frac{x}{x^2 - 4} = 2010.\]Clearing the denominators, we get
\begin{align*}
&x(x^2 - 2)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 3) \\
&\quad = 2010(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
\end{align*}This expands to
\[4x^7 + \dotsb = 2010x^8 + \dotsb,\]where only terms that have degree 7 or greater are shown.  Then
\[2010x^8 - 4x^7 + \dotsb = 0,\]so by Vieta's formulas, the sum of the roots is $\frac{4}{2010} = \boxed{\frac{2}{1005}}.$