Explanation:
Okay, so I need to solve this equation:

$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} + \frac{(x+3)(x-5)}{9(x+4)(x-6)} - \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = \frac{92}{585}
$$

Hmm, that looks pretty complicated. Let me try to break it down step by step. First, maybe I can simplify each term individually. Let me look at each fraction separately.

Starting with the first term: $\frac{(x+1)(x-3)}{5(x+2)(x-4)}$. Maybe I can expand the numerator and denominator? Let me try that.

Expanding the numerator: (x+1)(x-3) = x² - 3x + x - 3 = x² - 2x - 3.

Denominator: 5(x+2)(x-4) = 5[(x)(x-4) + 2(x-4)] = 5[x² - 4x + 2x - 8] = 5[x² - 2x - 8] = 5x² - 10x - 40.

Wait, but maybe expanding isn't the best approach here. Maybe there's a substitution or a pattern? Let me check the other terms.

Second term: $\frac{(x+3)(x-5)}{9(x+4)(x-6)}$. Let me expand numerator and denominator.

Numerator: (x+3)(x-5) = x² -5x + 3x -15 = x² - 2x -15.

Denominator: 9(x+4)(x-6) = 9[(x)(x-6) + 4(x-6)] = 9[x² -6x +4x -24] = 9[x² -2x -24] = 9x² - 18x - 216.

Third term: $- \frac{2(x+5)(x-7)}{13(x+6)(x-8)}$. Let me expand numerator and denominator.

Numerator: 2(x+5)(x-7) = 2[x² -7x +5x -35] = 2[x² -2x -35] = 2x² -4x -70.

Denominator: 13(x+6)(x-8) = 13[(x)(x-8) +6(x-8)] = 13[x² -8x +6x -48] = 13[x² -2x -48] = 13x² -26x -624.

So now, substituting back into the original equation:

$$
\frac{x² - 2x - 3}{5x² - 10x - 40} + \frac{x² - 2x -15}{9x² - 18x - 216} - \frac{2x² -4x -70}{13x² -26x -624} = \frac{92}{585}
$$

Hmm, looking at denominators, maybe I can factor them?

First denominator: 5x² -10x -40. Let me factor out a 5: 5(x² - 2x - 8). Then factor x² -2x -8: looking for two numbers that multiply to -8 and add to -2. That would be -4 and +2. So (x - 4)(x + 2). So denominator is 5(x - 4)(x + 2). Wait, that's the original denominator! So maybe expanding wasn't helpful. Let me check.

Original first term denominator: 5(x + 2)(x - 4). So same as above. So maybe I can write each term as:

First term: $\frac{(x+1)(x-3)}{5(x+2)(x-4)}$

Second term: $\frac{(x+3)(x-5)}{9(x+4)(x-6)}$

Third term: $- \frac{2(x+5)(x-7)}{13(x+6)(x-8)}$

So maybe there's a pattern here. Let me check the numerators and denominators.

Looking at the denominators:

First term denominator: (x + 2)(x - 4). Let me note that (x + 2) and (x - 4) are two terms apart? Wait, maybe not. Let me check the other denominators.

Second term denominator: (x + 4)(x - 6). Similarly, third term denominator: (x + 6)(x - 8). So each denominator is (x + 2 + 2k)(x - 4 - 2k) for k=0,1,2? Let me check:

For first term, k=0: (x + 2 + 0)(x - 4 - 0) = (x + 2)(x - 4). Correct.

Second term, k=1: (x + 2 + 2)(x - 4 - 2) = (x + 4)(x - 6). Correct.

Third term, k=2: (x + 2 + 4)(x - 4 - 4) = (x + 6)(x - 8). Correct. Interesting. So denominators are following a pattern. Similarly, numerators?

First numerator: (x + 1)(x - 3). Let me check if they follow a similar pattern. Let me think. If denominators are (x + 2 + 2k)(x - 4 - 2k), then maybe numerators are (x + 1 + 2k)(x - 3 - 2k)? Let me check for k=0: (x +1)(x -3). Correct. For k=1: (x + 3)(x -5). Which is the second numerator. For k=2: (x +5)(x -7). Which is the third numerator. So yes! So each term is of the form:

For k=0: $\frac{(x + 1 + 2k)(x - 3 - 2k)}{5(x + 2 + 2k)(x - 4 - 2k)}$

Similarly for k=1 and k=2, with denominators multiplied by 9 and 13 respectively. Wait, but the denominators have coefficients 5, 9, 13. Let me check:

First term denominator coefficient: 5. Second term: 9. Third term: 13. These are 5, 9, 13, which are increasing by 4 each time. So maybe there's a pattern here. So perhaps the general term is something like:

For k=0,1,2:

$$
\frac{(x + 1 + 2k)(x - 3 - 2k)}{(5 + 4k)(x + 2 + 2k)(x - 4 - 2k)}
$$

But the third term has a coefficient of -2 in front. Wait, original equation is:

First term: + [term], second term: + [term], third term: -2 [term]. So maybe the coefficients are 1, 1, -2? Not sure if that helps yet.

But maybe there's a substitution here. Let me think. Let me consider substituting y = x - something? Or maybe notice that denominators and numerators have similar structures. Let me think about the first term:

Let me write the first term as:

$$
\frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)} = \frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)}
$$

Let me try to write this as a function. Let me denote:

Let me think of each term as a function of (x + something). Let me check if there's a substitution that can make this easier. For example, maybe set t = x + something? Let me see.

Alternatively, maybe notice that each term can be expressed as a function of (x + 2) or something. Let me try to manipulate the first term:

Let me write the first term:

$$
\frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)} = \frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)}
$$

Let me try to write numerator in terms of denominator. Let me see:

Let me note that (x + 1) = (x + 2) - 1, and (x - 3) = (x - 4) + 1. So:

Numerator: [(x + 2) - 1][(x - 4) + 1] = (x + 2)(x - 4) + (x + 2) - (x - 4) - 1

Wait, expanding that:

Let me compute [(x + 2) - 1][(x - 4) + 1] = (x + 2)(x - 4) + (x + 2)(1) - 1*(x - 4) - 1*1

Which is (x + 2)(x - 4) + (x + 2) - (x - 4) - 1

Simplify:

(x + 2)(x - 4) + [ (x + 2) - (x - 4) ] - 1

Compute the bracket: (x + 2 - x + 4) = 6

So total numerator: (x + 2)(x - 4) + 6 - 1 = (x + 2)(x - 4) + 5

Therefore, the first term becomes:

$$
\frac{(x + 2)(x - 4) + 5}{5(x + 2)(x - 4)} = \frac{(x + 2)(x - 4)}{5(x + 2)(x - 4)} + \frac{5}{5(x + 2)(x - 4)} = \frac{1}{5} + \frac{1}{(x + 2)(x - 4)}
$$

Oh! That's a nice simplification! So the first term simplifies to 1/5 + 1/[(x + 2)(x - 4)]. Let me check that again.

Original numerator: (x + 1)(x - 3) = (x + 2 - 1)(x - 4 + 1) = (x + 2)(x - 4) + (x + 2) - (x - 4) - 1. Wait, earlier steps showed that equals (x + 2)(x - 4) + 5. So then dividing by denominator 5(x + 2)(x - 4) gives 1/5 + 1/[(x + 2)(x - 4)]. Yes, that seems correct. Let me verify with numbers. Let me pick x = 0.

Original first term: (0 + 1)(0 - 3)/[5(0 + 2)(0 - 4)] = (1)(-3)/[5(2)(-4)] = (-3)/(-40) = 3/40.

Simplified expression: 1/5 + 1/[(0 + 2)(0 - 4)] = 1/5 + 1/(2*(-4)) = 1/5 - 1/8 = (8 - 5)/40 = 3/40. Correct! So that works. Great!

So that simplification works. Let me try the same approach for the second term.

Second term: $\frac{(x + 3)(x - 5)}{9(x + 4)(x - 6)}$

Similarly, let me write numerator as [(x + 4) - 1][(x - 6) + 1]. Let me check:

(x + 3) = (x + 4) - 1, and (x - 5) = (x - 6) + 1. So:

Numerator: [(x + 4) - 1][(x - 6) + 1] = (x + 4)(x - 6) + (x + 4) - (x - 6) - 1

Expanding:

(x + 4)(x - 6) + [ (x + 4) - (x - 6) ] - 1

Compute the bracket: (x + 4 - x + 6) = 10

So numerator becomes: (x + 4)(x - 6) + 10 - 1 = (x + 4)(x - 6) + 9

Therefore, the second term is:

$$
\frac{(x + 4)(x - 6) + 9}{9(x + 4)(x - 6)} = \frac{(x + 4)(x - 6)}{9(x + 4)(x - 6)} + \frac{9}{9(x + 4)(x - 6)} = \frac{1}{9} + \frac{1}{(x + 4)(x - 6)}
$$

Again, let me check with x = 0:

Original second term: (0 + 3)(0 - 5)/[9(0 + 4)(0 - 6)] = (3)(-5)/[9*4*(-6)] = (-15)/(-216) = 5/72.

Simplified expression: 1/9 + 1/[(0 + 4)(0 - 6)] = 1/9 + 1/(4*(-6)) = 1/9 - 1/24 = (8 - 3)/72 = 5/72. Correct! Nice.

Third term: $- \frac{2(x + 5)(x - 7)}{13(x + 6)(x - 8)}$

Let me try the same approach. Let me write numerator as [(x + 6) - 1][(x - 8) + 1]. Let me check:

(x + 5) = (x + 6) - 1, and (x - 7) = (x - 8) + 1. So:

Numerator: [(x + 6) - 1][(x - 8) + 1] = (x + 6)(x - 8) + (x + 6) - (x - 8) - 1

Expanding:

(x + 6)(x - 8) + [ (x + 6) - (x - 8) ] - 1

Compute the bracket: (x + 6 - x + 8) = 14

So numerator becomes: (x + 6)(x - 8) + 14 - 1 = (x + 6)(x - 8) + 13

Therefore, the third term is:

$$
- \frac{2[(x + 6)(x - 8) + 13]}{13(x + 6)(x - 8)} = - \left[ \frac{2(x + 6)(x - 8)}{13(x + 6)(x - 8)} + \frac{2*13}{13(x + 6)(x - 8)} \right] = - \left[ \frac{2}{13} + \frac{2}{(x + 6)(x - 8)} \right] = -\frac{2}{13} - \frac{2}{(x + 6)(x - 8)}
$$

Let me check with x = 0 again:

Original third term: -2*(0 + 5)(0 - 7)/[13*(0 + 6)(0 - 8)] = -2*(5)(-7)/[13*6*(-8)] = -2*(-35)/[13*(-48)] = 70 / (-624) = -35/312.

Simplified expression: -2/13 - 2/[(0 + 6)(0 - 8)] = -2/13 - 2/(6*(-8)) = -2/13 - 2/(-48) = -2/13 + 1/24. Let me compute that:

Convert to common denominator, which is 312. -2/13 = -48/312, 1/24 = 13/312. So total is (-48 + 13)/312 = -35/312. Correct! Perfect.

So now, substituting all three terms back into the original equation:

First term: 1/5 + 1/[(x + 2)(x - 4)]

Second term: 1/9 + 1/[(x + 4)(x - 6)]

Third term: -2/13 - 2/[(x + 6)(x - 8)]

So the entire equation becomes:

$$
\left( \frac{1}{5} + \frac{1}{(x + 2)(x - 4)} \right) + \left( \frac{1}{9} + \frac{1}{(x + 4)(x - 6)} \right) - \left( \frac{2}{13} + \frac{2}{(x + 6)(x - 8)} \right) = \frac{92}{585}
$$

Let me simplify the constants first. Let me compute 1/5 + 1/9 - 2/13.

Compute 1/5 + 1/9: common denominator is 45. 9/45 + 5/45 = 14/45.

Then subtract 2/13: 14/45 - 2/13. Common denominator is 585 (since 45*13=585). So:

14/45 = (14*13)/585 = 182/585

2/13 = (2*45)/585 = 90/585

So 182/585 - 90/585 = 92/585.

Wait! That's exactly the right-hand side of the equation. So the constants add up to 92/585. Therefore, the equation simplifies to:

$$
\frac{92}{585} + \frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = \frac{92}{585}
$$

Subtracting 92/585 from both sides:

$$
\frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = 0
$$

So now, the equation reduces to:

$$
\frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = 0
$$

This seems more manageable. Let me denote each term as follows:

Let me call A = 1/[(x + 2)(x - 4)], B = 1/[(x + 4)(x - 6)], C = 2/[(x + 6)(x - 8)]. So equation is A + B - C = 0.

But maybe I can combine these terms. Let me try to combine them step by step.

First, let me combine A and B. Let me find a common denominator for A and B. The denominators are (x + 2)(x - 4) and (x + 4)(x - 6). The common denominator would be (x + 2)(x - 4)(x + 4)(x - 6). But that might be complicated. Alternatively, maybe there's a pattern here.

Alternatively, notice that each denominator is similar to the previous one, shifted by 2 in x. Let me check:

First denominator: (x + 2)(x - 4). Let me write it as (x + 2)(x - 4) = (x + 2)(x - 4)

Second denominator: (x + 4)(x - 6) = (x + 4)(x - 6)

Third denominator: (x + 6)(x - 8) = (x + 6)(x - 8)

So each denominator is (x + 2 + 2k)(x - 4 - 2k) for k=0,1,2. So similar to before.

Alternatively, maybe I can write each term as a difference of fractions. Let me think. For example, similar to how we did before with the first term.

Let me try to express each term as a difference. Let me take the first term: 1/[(x + 2)(x - 4)]. Let me try to write this as something like [A/(x + 2) - B/(x - 4)].

Let me suppose that:

1/[(x + 2)(x - 4)] = A/(x + 2) + B/(x - 4). Wait, but if I want to write it as a difference, maybe:

Let me think of partial fractions. Let me recall that:

1/[(x + a)(x + b)] = [1/(a - b)] [1/(x + b) - 1/(x + a)]

Wait, let me check:

Let me suppose that 1/[(x + a)(x + b)] = C/(x + a) + D/(x + b). Then:

1 = C(x + b) + D(x + a). Let me set x = -a: 1 = C(0) + D(-a + a) = D(0)? Wait, no. Wait, if x = -a, then:

1 = C( -a + b ) + D(0) => C = 1/(b - a). Similarly, x = -b: 1 = C(0) + D(-b + a) => D = 1/(a - b). Therefore:

1/[(x + a)(x + b)] = [1/(b - a)] [1/(x + a) - 1/(x + b)]

So in our case, for the first term, denominator is (x + 2)(x - 4). Let me write it as (x - (-2))(x - 4). So a = -2, b = 4. Then:

1/[(x + 2)(x - 4)] = [1/(4 - (-2))] [1/(x + 2) - 1/(x - 4)] = [1/6][1/(x + 2) - 1/(x - 4)]

Similarly for the second term: 1/[(x + 4)(x - 6)] = [1/( -6 - 4 )][1/(x + 4) - 1/(x - 6)]? Wait, let me check:

Wait, using the formula above, with a = -4, b = 6. Then:

1/[(x + 4)(x - 6)] = [1/(6 - (-4))][1/(x + 4) - 1/(x - 6)] = [1/10][1/(x + 4) - 1/(x - 6)]

Similarly, third term: 2/[(x + 6)(x - 8)] = 2 * [1/( -8 - 6 )][1/(x + 6) - 1/(x - 8)] = 2 * [1/(-14)][1/(x + 6) - 1/(x - 8)] = -2/14 [1/(x + 6) - 1/(x - 8)] = -1/7 [1/(x + 6) - 1/(x - 8)]

So substituting back into the equation:

First term: [1/6][1/(x + 2) - 1/(x - 4)]

Second term: [1/10][1/(x + 4) - 1/(x - 6)]

Third term: - [1/7][1/(x + 6) - 1/(x - 8)]

So the equation becomes:

[1/6][1/(x + 2) - 1/(x - 4)] + [1/10][1/(x + 4) - 1/(x - 6)] - [1/7][1/(x + 6) - 1/(x - 8)] = 0

Let me write this out:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) - (1/7)(1/(x + 6) - 1/(x - 8)) = 0

This seems complicated, but maybe there's a telescoping pattern here. Let me expand each term:

First term: (1/6)(1/(x + 2)) - (1/6)(1/(x - 4))

Second term: (1/10)(1/(x + 4)) - (1/10)(1/(x - 6))

Third term: - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8))

So combining all terms:

[ (1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) ] + [ (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) ] + [ - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8)) ] = 0

Now, let me rearrange the terms by their denominators:

Terms with 1/(x + 2): (1/6)(1/(x + 2))

Terms with 1/(x + 4): (1/10)(1/(x + 4))

Terms with 1/(x - 4): - (1/6)(1/(x - 4))

Terms with 1/(x - 6): - (1/10)(1/(x - 6))

Terms with 1/(x + 6): - (1/7)(1/(x + 6))

Terms with 1/(x - 8): + (1/7)(1/(x - 8))

So, maybe there's a pattern here. Let me check if there's a telescoping effect. Let me think about the denominators:

Looking at denominators: x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. It seems like each term is related to the previous one by adding or subtracting 2. Maybe if I can pair terms?

Alternatively, maybe I can combine terms with similar denominators. Let me see.

Alternatively, maybe I can consider that the equation is a combination of terms that can be grouped. Let me think.

Alternatively, maybe I can write all terms with denominators as follows:

Let me note that denominators are x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. Let me think of them as x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. Maybe if I consider variables shifted by some amount?

Alternatively, maybe I can consider that the equation is symmetric in some way? Not sure. Alternatively, maybe I can combine terms step by step.

Let me try to combine the first two terms:

First, let me consider the terms with 1/(x + 2) and 1/(x + 4). But they are separate. Maybe not helpful.

Alternatively, let me think of the entire expression as:

Let me denote:

Let me write all terms:

Term1: (1/6)(1/(x + 2))

Term2: (1/10)(1/(x + 4))

Term3: - (1/6)(1/(x - 4))

Term4: - (1/10)(1/(x - 6))

Term5: - (1/7)(1/(x + 6))

Term6: + (1/7)(1/(x - 8))

So maybe I can group Term1 and Term3, Term2 and Term4, Term5 and Term6?

Let me try:

Group Term1 and Term3: (1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) = (1/6)[1/(x + 2) - 1/(x - 4)]

Similarly, Group Term2 and Term4: (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) = (1/10)[1/(x + 4) - 1/(x - 6)]

Group Term5 and Term6: - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8)) = (1/7)[ -1/(x + 6) + 1/(x - 8) ] = (1/7)[1/(x - 8) - 1/(x + 6)]

Wait, but this brings us back to the original expression. So that doesn't help. Maybe another grouping?

Alternatively, maybe consider that denominators are spaced by 2, so maybe if I consider variables shifted by 2?

Alternatively, maybe I can consider substituting y = x - something. Let me think. Let me check if there's a substitution that can make denominators symmetric.

Alternatively, maybe I can combine all terms over a common denominator. But that might be very complicated. Let me think.

Alternatively, maybe I can notice that denominators are (x + 2), (x + 4), (x - 4), (x - 6), (x + 6), (x - 8). Let me note that (x - 4) = (x + 2) - 6, but not sure.

Alternatively, maybe I can consider that denominators are related to each other. For example, (x + 2) and (x - 4) differ by 6, (x + 4) and (x - 6) differ by 10, etc. Not sure.

Alternatively, maybe I can consider that the entire equation is a combination of terms that can be telescoped when considering more terms. Wait, but there are only three terms. Let me think again.

Alternatively, maybe I can consider that the equation is similar to a telescoping series. Let me think of the terms as differences. For example, if I have terms like 1/(x + 2) - 1/(x - 4), etc., but with coefficients.

Alternatively, maybe I can write the entire equation as:

Let me think of the entire expression as:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Wait, that's the same as before. So maybe I can write this as:

Let me denote:

Let me think of each bracket as a difference. Let me write:

Let me consider the first bracket: 1/(x + 2) - 1/(x - 4)

Second bracket: 1/(x + 4) - 1/(x - 6)

Third bracket: 1/(x - 8) - 1/(x + 6)

But with coefficients 1/6, 1/10, 1/7 respectively.

But how can I combine these? Maybe if I can find a common pattern or relation between these brackets?

Alternatively, maybe I can consider that each bracket is similar to the previous one, shifted by 2. Let me check:

First bracket: 1/(x + 2) - 1/(x - 4)

Second bracket: 1/(x + 4) - 1/(x - 6)

Third bracket: 1/(x - 8) - 1/(x + 6)

If I look at the first bracket, if I shift x by 2, I get the second bracket. Similarly, shifting x by 2 again, I get something else. But not sure.

Alternatively, maybe I can consider that the entire expression is a combination of terms that can be telescoped when considering more terms. But since there are only three terms, maybe not.

Alternatively, maybe I can consider that the equation is equal to zero, so:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Let me write all terms:

(1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) + (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) - (1/7)(1/(x + 6)) = 0

Now, let me rearrange the terms:

[ (1/6)(1/(x + 2)) + (1/10)(1/(x + 4)) - (1/7)(1/(x + 6)) ] + [ - (1/6)(1/(x - 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) ] = 0

Hmm, not sure. Maybe I can consider that the terms with positive signs and negative signs are separate. Let me think.

Alternatively, maybe I can consider that denominators are x + 2, x + 4, x + 6, x - 4, x - 6, x - 8. Let me note that x - 4 = (x + 2) - 6, x - 6 = (x + 4) - 10, x - 8 = (x + 6) - 14. Not sure.

Alternatively, maybe I can make a substitution. Let me set y = x - something. Let me think. For example, if I set y = x - 3, maybe? Let me try.

But maybe not. Alternatively, maybe set t = x - something to make denominators symmetric. Let me think.

Alternatively, maybe I can consider that denominators are (x + 2), (x + 4), (x + 6), and their counterparts (x - 4), (x - 6), (x - 8). Let me note that (x - 4) = (x + 2) - 6, (x - 6) = (x + 4) - 10, (x - 8) = (x + 6) - 14. Not sure.

Alternatively, maybe I can consider that denominators are in arithmetic progression? For example, x + 2, x + 4, x + 6, etc. But the other denominators are x - 4, x - 6, x - 8. So maybe if I consider two sequences: one increasing by 2, and one decreasing by 2.

Alternatively, maybe I can think of the entire equation as a combination of terms that can be expressed in terms of 1/(x + 2), 1/(x + 4), 1/(x + 6), etc. Let me try to write all terms in terms of these.

Let me note that:

- (1/6)(1/(x - 4)) = - (1/6)(1/(x + 2 - 6)) = - (1/6)(1/(x + 2 - 6))

Similarly, - (1/10)(1/(x - 6)) = - (1/10)(1/(x + 4 - 10))

And + (1/7)(1/(x - 8)) = + (1/7)(1/(x + 6 - 14))

But this seems too vague. Maybe not helpful.

Alternatively, maybe I can consider that the equation is a linear combination of terms like 1/(x + 2), 1/(x + 4), 1/(x + 6), 1/(x - 4), 1/(x - 6), 1/(x - 8). Let me denote:

Let me write the equation as:

(1/6)(1/(x + 2)) + (1/10)(1/(x + 4)) - (1/7)(1/(x + 6)) - (1/6)(1/(x - 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) = 0

Let me group terms with similar denominators:

For 1/(x + 2): (1/6)

For 1/(x + 4): (1/10)

For 1/(x + 6): - (1/7)

For 1/(x - 4): - (1/6)

For 1/(x - 6): - (1/10)

For 1/(x - 8): + (1/7)

So maybe I can write this as:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Which is the same as before. So I need another approach.

Alternatively, maybe I can consider that the equation is equal to zero, so:

Let me denote:

Let me consider that denominators are (x + 2), (x + 4), (x + 6), (x - 4), (x - 6), (x - 8). Let me think of them as pairs:

(x + 2) and (x - 4), (x + 4) and (x - 6), (x + 6) and (x - 8). Each pair differs by 6, 10, 14? Wait, (x + 2) and (x - 4): difference is 6. Similarly, (x + 4) and (x - 6): difference is 10. (x + 6) and (x - 8): difference is 14. So differences are 6, 10, 14. Which are increasing by 4 each time. Interesting. So maybe there's a pattern here.

But how does that help? Maybe not directly.

Alternatively, maybe I can consider that each pair is related to the previous one. For example, the first pair (x + 2) and (x - 4) can be thought of as (x + 2) and (x - 4), then the next pair is (x + 4) and (x - 6), which is similar but shifted by 2. Maybe if I consider that the entire equation is a telescoping sum?

Alternatively, maybe I can consider that the equation is a combination of terms that can be expressed as differences of fractions with denominators that telescope when combined. For example, if I can write the entire expression as a single fraction, maybe it simplifies.

But combining all terms would be very tedious. Let me try.

First, let me note that denominators are:

For the first term: (x + 2)(x - 4)

Second term: (x + 4)(x - 6)

Third term: (x + 6)(x - 8)

So the common denominator would be the product of all three denominators: (x + 2)(x - 4)(x + 4)(x - 6)(x + 6)(x - 8). That's a huge denominator. But maybe manageable?

Alternatively, maybe I can combine two terms at a time. Let me try combining the first two terms first.

Let me take the first two terms:

Term1 + Term2 = [1/(x + 2)(x - 4)] + [1/(x + 4)(x - 6)]

Let me find a common denominator for these two terms: (x + 2)(x - 4)(x + 4)(x - 6). Let me compute:

Term1: [ (x + 4)(x - 6) ] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] + [ (x + 2)(x - 4) ] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ]

So numerator: (x + 4)(x - 6) + (x + 2)(x - 4)

Let me compute that:

First part: (x + 4)(x - 6) = x² - 6x + 4x - 24 = x² - 2x - 24

Second part: (x + 2)(x - 4) = x² - 4x + 2x - 8 = x² - 2x - 8

Adding them together: (x² - 2x - 24) + (x² - 2x - 8) = 2x² - 4x - 32

Therefore, Term1 + Term2 = [2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ]

Simplify numerator: 2x² - 4x - 32 = 2(x² - 2x - 16). Not sure if factors. Let me check discriminant: 4 + 64 = 68. Not a perfect square. So maybe leave as is.

Now, adding Term3: - [2 / (x + 6)(x - 8)]

So the entire equation is:

[2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] - [2 / (x + 6)(x - 8)] = 0

Let me write this as:

[2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] = [2 / (x + 6)(x - 8)]

Multiply both sides by (x + 2)(x - 4)(x + 4)(x - 6)(x + 6)(x - 8) to eliminate denominators:

Left side: [2x² - 4x - 32] * (x + 6)(x - 8)

Right side: 2 * (x + 2)(x - 4)(x + 4)(x - 6)

So equation becomes:

(2x² - 4x - 32)(x + 6)(x - 8) = 2(x + 2)(x - 4)(x + 4)(x - 6)

Let me simplify both sides.

First, factor out 2 from the left side:

Left side: 2(x² - 2x - 16)(x + 6)(x - 8)

Right side: 2(x + 2)(x - 4)(x + 4)(x - 6)

We can divide both sides by 2:

(x² - 2x - 16)(x + 6)(x - 8) = (x + 2)(x - 4)(x + 4)(x - 6)

Now, let me compute both sides.

First, let me compute the right side: (x + 2)(x - 4)(x + 4)(x - 6)

Note that (x + 2)(x - 4) = x² - 2x - 8 and (x + 4)(x - 6) = x² - 2x - 24. So the right side is (x² - 2x - 8)(x² - 2x - 24). Let me denote y = x² - 2x. Then right side becomes (y - 8)(y - 24) = y² - 32y + 192.

Similarly, left side: (x² - 2x - 16)(x + 6)(x - 8). Let me compute (x + 6)(x - 8) = x² - 2x - 48. So left side is (x² - 2x - 16)(x² - 2x - 48). Let me denote y = x² - 2x again. Then left side becomes (y - 16)(y - 48) = y² - 64y + 768.

So now, equation becomes:

Left side: y² - 64y + 768 = Right side: y² - 32y + 192

Subtracting right side from both sides:

(y² - 64y + 768) - (y² - 32y + 192) = 0 => -32y + 576 = 0 => -32y + 576 = 0 => 32y = 576 => y = 576 / 32 = 18.

So y = 18. Recall that y = x² - 2x. Therefore:

x² - 2x = 18 => x² - 2x - 18 = 0.

Solving this quadratic equation:

x = [2 ± sqrt(4 + 72)] / 2 = [2 ± sqrt(76)] / 2 = [2 ± 2*sqrt(19)] / 2 = 1 ± sqrt(19).

So the solutions are x = 1 + sqrt(19) and x = 1 - sqrt(19).

But wait, I need to check if these solutions are valid, i.e., they don't make any denominators zero in the original equation.

Original equation denominators are:

First term: 5(x + 2)(x - 4)

Second term: 9(x + 4)(x - 6)

Third term: 13(x + 6)(x - 8)

So we need to ensure that x ≠ -2, 4, -4, 6, -6, 8.

Let me check for x = 1 + sqrt(19) and x = 1 - sqrt(19).

First, approximate sqrt(19) ≈ 4.3589.

So x ≈ 1 + 4.3589 ≈ 5.3589, and x ≈ 1 - 4.3589 ≈ -3.3589.

Check if these values are equal to -2, 4, -4, 6, -6, 8.

For x ≈ 5.3589: not equal to any of those. For x ≈ -3.3589: not equal to any of those. So both solutions are valid.

Therefore, the real roots are x = 1 + sqrt(19) and x = 1 - sqrt(19).

But let me verify this solution by plugging back into the original equation to ensure no mistakes were made during the algebra.

Let me check with x = 1 + sqrt(19). Let me compute the left-hand side (LHS) and see if it equals 92/585.

But this might be tedious. Alternatively, since we followed the algebra correctly, and the steps were reversible (except for multiplying both sides by denominators, which we need to check for extraneous solutions, but we already checked that the solutions don't make denominators zero), so likely correct.

But let me check with x = 1 + sqrt(19). Let me compute y = x² - 2x = 18. So that was the key step. Since we derived that from the equation, and the rest of the steps were algebraic manipulations, it should be correct.

Therefore, the real roots are 1 ± sqrt(19).

But let me check if there are any other solutions. Wait, when we divided both sides by 2, and then simplified, we arrived at y = 18, leading to quadratic equation. So only two solutions. Therefore, the real roots are 1 + sqrt(19) and 1 - sqrt(19).

But let me check if there was any step where we might have lost solutions or introduced extraneous ones. For example, when we multiplied both sides by denominators, we assumed that denominators are not zero, which we checked. Also, when we divided by 2, that's fine. So I think these are the only solutions.

Therefore, the real roots are 1 + sqrt(19) and 1 - sqrt(19).

But let me check if the original equation simplifies correctly. Let me take x = 1 + sqrt(19). Let me compute the left-hand side.

But maybe it's too time-consuming. Alternatively, since we followed the steps correctly, and the algebra led us to a quadratic equation with two solutions, and they don't make denominators zero, they should be valid.

Therefore, the answer is x = 1 ± sqrt(19). So in boxed form, they should be written as \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

But the problem says "Enter the real roots, separated by commas." So likely need to write both in one box, separated by commas.

**Final Answer**
The real roots are \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

Wait, but the instructions say "Enter the real roots, separated by commas." So maybe inside a single box? Let me check the problem statement again.

"Enter the real roots, separated by commas."

So likely, they want both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system expects multiple boxed answers? Or maybe a single box with both. Let me check standard practice. Usually, for multiple answers, they are separated by commas inside a single box. For example, \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} might be two separate boxes, but if the system allows only one box, then inside one box. Since the original problem says "Enter the real roots, separated by commas," I think they expect both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system requires a single box. Let me check the initial problem statement. The user says "put your final answer within \boxed{}". So maybe they expect multiple boxed answers? But the standard practice here is to have one box. Maybe the answer is two numbers, so written as \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

But since the user might expect comma-separated inside one box, I'll check similar problems. For example, if there are multiple solutions, they are often listed in one box separated by commas. So I think:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system requires a single box. Let me check the original problem's example. The original equation is given, and the user says "Enter the real roots, separated by commas." So likely, they want both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But actually, in many cases, for multiple answers, they are put in one box. For example, on AoPS or other platforms, multiple answers are separated by commas in a single box. So I think the correct format is:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since the user might have specific instructions, but given the initial problem says "put your final answer within \boxed{}", maybe they expect multiple boxed answers. However, since the user hasn't specified, but in the problem statement says "Enter the real roots, separated by commas," I think safest is to put both in one box separated by commas.

**Final Answer**
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

Wait, but actually, looking at the problem again, the original equation is equal to 92/585. When I simplified, I found that the constants added up to 92/585, so the equation reduced to the sum of the fractions equals zero. Then through substitutions, I arrived at the quadratic equation. So the solutions are correct. Therefore, the real roots are 1 ± sqrt(19). So I think the answer is those two values. So I'll go with:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But if the system requires a single box, maybe:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since the user might have intended multiple boxed answers, but the instruction says "put your final answer within \boxed{}", which is singular. Maybe they expect both answers in one box. For example:

\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

Yes, that's possible. Let me check standard practice. In many cases, for multiple answers, they are put in one box separated by commas. For example, on AoPS or math competitions, they might write like that. So I think that's the way to go.

**Final Answer**
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

Wait, but the system might not accept multiple boxes. Since the user hasn't specified, but given the initial problem says "Enter the real roots, separated by commas," I think the correct format is:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since I can't be sure, but given the original instruction says "put your final answer within \boxed{}", maybe they expect both answers in one box. So:

\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

Yes, that's likely. So I'll go with that.

**Final Answer**
\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

To solve the equation:

$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} + \frac{(x+3)(x-5)}{9(x+4)(x-6)} - \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = \frac{92}{585}
$$

we begin by simplifying each term using a clever algebraic technique. Each term has a numerator and denominator that are products of linear expressions, and we observe that the numerators and denominators are related by a consistent pattern.

---

### Step 1: Simplify Each Term

We use the identity:

$$
\frac{(x + a)(x - b)}{(x + c)(x - d)} = \frac{1}{(c - d)} \left( \frac{1}{x + c} - \frac{1}{x - d} \right)
$$

This allows us to rewrite each term as a difference of two simpler fractions.

#### First Term:
$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} = \frac{1}{5} + \frac{1}{(x+2)(x-4)}
$$

#### Second Term:
$$
\frac{(x+3)(x-5)}{9(x+4)(x-6)} = \frac{1}{9} + \frac{1}{(x+4)(x-6)}
$$

#### Third Term:
$$
- \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = -\frac{2}{13} - \frac{2}{(x+6)(x-8)}
$$

---

### Step 2: Combine All Terms

Substituting back into the original equation:

$$
\left( \frac{1}{5} + \frac{1}{(x+2)(x-4)} \right) + \left( \frac{1}{9} + \frac{1}{(x+4)(x-6)} \right) - \left( \frac{2}{13} + \frac{2}{(x+6)(x-8)} \right) = \frac{92}{585}
$$

Now, combine the constants:

$$
\frac{1}{5} + \frac{1}{9} - \frac{2}{13} = \frac{92}{585}
$$

This confirms that the constants on the left-hand side match the right-hand side. So the equation reduces to:

$$
\frac{1}{(x+2)(x-4)} + \frac{1}{(x+4)(x-6)} - \frac{2}{(x+6)(x-8)} = 0
$$

---

### Step 3: Use Partial Fractions

We apply the identity again to each term:

- $\frac{1}{(x+2)(x-4)} = \frac{1}{6} \left( \frac{1}{x+2} - \frac{1}{x-4} \right)$
- $\frac{1}{(x+4)(x-6)} = \frac{1}{10} \left( \frac{1}{x+4} - \frac{1}{x-6} \right)$
- $\frac{2}{(x+6)(x-8)} = \frac{2}{14} \left( \frac{1}{x+6} - \frac{1}{x-8} \right) = \frac{1}{7} \left( \frac{1}{x+6} - \frac{1}{x-8} \right)$

Substituting back:

$$
\frac{1}{6} \left( \frac{1}{x+2} - \frac{1}{x-4} \right) + \frac{1}{10} \left( \frac{1}{x+4} - \frac{1}{x-6} \right) - \frac{1}{7} \left( \frac{1}{x+6} - \frac{1}{x-8} \right) = 0
$$

---

### Step 4: Combine All Terms

Grouping all terms:

$$
\frac{1}{6} \cdot \frac{1}{x+2} - \frac{1}{6} \cdot \frac{1}{x-4} + \frac{1}{10} \cdot \frac{1}{x+4} - \frac{1}{10} \cdot \frac{1}{x-6} - \frac{1}{7} \cdot \frac{1}{x+6} + \frac{1}{7} \cdot \frac{1}{x-8} = 0
$$

Now, we combine all terms over a common denominator. However, instead of expanding, we observe that the entire expression simplifies to a quadratic equation in terms of $x^2 - 2x$.

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

$$
(y - 16)(y - 48) = (y - 8)(y - 24)
$$

Expanding both sides:

- Left: $y^2 - 64y + 768$
- Right: $y^2 - 32y + 192$

Subtracting:

$$
(y^2 - 64y + 768) - (y^2 - 32y + 192) = -32y + 576 = 0
\Rightarrow y = 18
$$

Recall $y = x^2 - 2x$, so:

$$
x^2 - 2x = 18 \Rightarrow x^2 - 2x - 18 = 0
$$

Solving this quadratic:

$$
x = \frac{2 \pm \sqrt{4 + 72}}{2} = \frac{2 \pm \sqrt{76}}{2} = 1 \pm \sqrt{19}
$$

---

### Step 5: Final Answer

We check that these values do not make any denominator zero in the original equation. Since the denominators are:

- $x + 2$, $x - 4$, $x + 4$, $x - 6$, $x + 6$, $x - 8$

and $1 \pm \sqrt{19} \approx 5.3589$ and $-3.3589$, none of these values are equal to any of the excluded values, so both are valid.

---

### ✅ Final Answer:

$$
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation:

$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} + \frac{(x+3)(x-5)}{9(x+4)(x-6)} - \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = \frac{92}{585}
$$

Hmm, that looks pretty complicated. Let me try to break it down step by step. First, maybe I can simplify each term individually. Let me look at each fraction separately.

Starting with the first term: $\frac{(x+1)(x-3)}{5(x+2)(x-4)}$. Maybe I can expand the numerator and denominator? Let me try that.

Expanding the numerator: (x+1)(x-3) = x² - 3x + x - 3 = x² - 2x - 3.

Denominator: 5(x+2)(x-4) = 5[(x)(x-4) + 2(x-4)] = 5[x² - 4x + 2x - 8] = 5[x² - 2x - 8] = 5x² - 10x - 40.

Wait, but maybe expanding isn't the best approach here. Maybe there's a substitution or a pattern? Let me check the other terms.

Second term: $\frac{(x+3)(x-5)}{9(x+4)(x-6)}$. Let me expand numerator and denominator.

Numerator: (x+3)(x-5) = x² -5x + 3x -15 = x² - 2x -15.

Denominator: 9(x+4)(x-6) = 9[(x)(x-6) + 4(x-6)] = 9[x² -6x +4x -24] = 9[x² -2x -24] = 9x² - 18x - 216.

Third term: $- \frac{2(x+5)(x-7)}{13(x+6)(x-8)}$. Let me expand numerator and denominator.

Numerator: 2(x+5)(x-7) = 2[x² -7x +5x -35] = 2[x² -2x -35] = 2x² -4x -70.

Denominator: 13(x+6)(x-8) = 13[(x)(x-8) +6(x-8)] = 13[x² -8x +6x -48] = 13[x² -2x -48] = 13x² -26x -624.

So now, substituting back into the original equation:

$$
\frac{x² - 2x - 3}{5x² - 10x - 40} + \frac{x² - 2x -15}{9x² - 18x - 216} - \frac{2x² -4x -70}{13x² -26x -624} = \frac{92}{585}
$$

Hmm, looking at denominators, maybe I can factor them?

First denominator: 5x² -10x -40. Let me factor out a 5: 5(x² - 2x - 8). Then factor x² -2x -8: looking for two numbers that multiply to -8 and add to -2. That would be -4 and +2. So (x - 4)(x + 2). So denominator is 5(x - 4)(x + 2). Wait, that's the original denominator! So maybe expanding wasn't helpful. Let me check.

Original first term denominator: 5(x + 2)(x - 4). So same as above. So maybe I can write each term as:

First term: $\frac{(x+1)(x-3)}{5(x+2)(x-4)}$

Second term: $\frac{(x+3)(x-5)}{9(x+4)(x-6)}$

Third term: $- \frac{2(x+5)(x-7)}{13(x+6)(x-8)}$

So maybe there's a pattern here. Let me check the numerators and denominators.

Looking at the denominators:

First term denominator: (x + 2)(x - 4). Let me note that (x + 2) and (x - 4) are two terms apart? Wait, maybe not. Let me check the other denominators.

Second term denominator: (x + 4)(x - 6). Similarly, third term denominator: (x + 6)(x - 8). So each denominator is (x + 2 + 2k)(x - 4 - 2k) for k=0,1,2? Let me check:

For first term, k=0: (x + 2 + 0)(x - 4 - 0) = (x + 2)(x - 4). Correct.

Second term, k=1: (x + 2 + 2)(x - 4 - 2) = (x + 4)(x - 6). Correct.

Third term, k=2: (x + 2 + 4)(x - 4 - 4) = (x + 6)(x - 8). Correct. Interesting. So denominators are following a pattern. Similarly, numerators?

First numerator: (x + 1)(x - 3). Let me check if they follow a similar pattern. Let me think. If denominators are (x + 2 + 2k)(x - 4 - 2k), then maybe numerators are (x + 1 + 2k)(x - 3 - 2k)? Let me check for k=0: (x +1)(x -3). Correct. For k=1: (x + 3)(x -5). Which is the second numerator. For k=2: (x +5)(x -7). Which is the third numerator. So yes! So each term is of the form:

For k=0: $\frac{(x + 1 + 2k)(x - 3 - 2k)}{5(x + 2 + 2k)(x - 4 - 2k)}$

Similarly for k=1 and k=2, with denominators multiplied by 9 and 13 respectively. Wait, but the denominators have coefficients 5, 9, 13. Let me check:

First term denominator coefficient: 5. Second term: 9. Third term: 13. These are 5, 9, 13, which are increasing by 4 each time. So maybe there's a pattern here. So perhaps the general term is something like:

For k=0,1,2:

$$
\frac{(x + 1 + 2k)(x - 3 - 2k)}{(5 + 4k)(x + 2 + 2k)(x - 4 - 2k)}
$$

But the third term has a coefficient of -2 in front. Wait, original equation is:

First term: + [term], second term: + [term], third term: -2 [term]. So maybe the coefficients are 1, 1, -2? Not sure if that helps yet.

But maybe there's a substitution here. Let me think. Let me consider substituting y = x - something? Or maybe notice that denominators and numerators have similar structures. Let me think about the first term:

Let me write the first term as:

$$
\frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)} = \frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)}
$$

Let me try to write this as a function. Let me denote:

Let me think of each term as a function of (x + something). Let me check if there's a substitution that can make this easier. For example, maybe set t = x + something? Let me see.

Alternatively, maybe notice that each term can be expressed as a function of (x + 2) or something. Let me try to manipulate the first term:

Let me write the first term:

$$
\frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)} = \frac{(x + 1)(x - 3)}{5(x + 2)(x - 4)}
$$

Let me try to write numerator in terms of denominator. Let me see:

Let me note that (x + 1) = (x + 2) - 1, and (x - 3) = (x - 4) + 1. So:

Numerator: [(x + 2) - 1][(x - 4) + 1] = (x + 2)(x - 4) + (x + 2) - (x - 4) - 1

Wait, expanding that:

Let me compute [(x + 2) - 1][(x - 4) + 1] = (x + 2)(x - 4) + (x + 2)(1) - 1*(x - 4) - 1*1

Which is (x + 2)(x - 4) + (x + 2) - (x - 4) - 1

Simplify:

(x + 2)(x - 4) + [ (x + 2) - (x - 4) ] - 1

Compute the bracket: (x + 2 - x + 4) = 6

So total numerator: (x + 2)(x - 4) + 6 - 1 = (x + 2)(x - 4) + 5

Therefore, the first term becomes:

$$
\frac{(x + 2)(x - 4) + 5}{5(x + 2)(x - 4)} = \frac{(x + 2)(x - 4)}{5(x + 2)(x - 4)} + \frac{5}{5(x + 2)(x - 4)} = \frac{1}{5} + \frac{1}{(x + 2)(x - 4)}
$$

Oh! That's a nice simplification! So the first term simplifies to 1/5 + 1/[(x + 2)(x - 4)]. Let me check that again.

Original numerator: (x + 1)(x - 3) = (x + 2 - 1)(x - 4 + 1) = (x + 2)(x - 4) + (x + 2) - (x - 4) - 1. Wait, earlier steps showed that equals (x + 2)(x - 4) + 5. So then dividing by denominator 5(x + 2)(x - 4) gives 1/5 + 1/[(x + 2)(x - 4)]. Yes, that seems correct. Let me verify with numbers. Let me pick x = 0.

Original first term: (0 + 1)(0 - 3)/[5(0 + 2)(0 - 4)] = (1)(-3)/[5(2)(-4)] = (-3)/(-40) = 3/40.

Simplified expression: 1/5 + 1/[(0 + 2)(0 - 4)] = 1/5 + 1/(2*(-4)) = 1/5 - 1/8 = (8 - 5)/40 = 3/40. Correct! So that works. Great!

So that simplification works. Let me try the same approach for the second term.

Second term: $\frac{(x + 3)(x - 5)}{9(x + 4)(x - 6)}$

Similarly, let me write numerator as [(x + 4) - 1][(x - 6) + 1]. Let me check:

(x + 3) = (x + 4) - 1, and (x - 5) = (x - 6) + 1. So:

Numerator: [(x + 4) - 1][(x - 6) + 1] = (x + 4)(x - 6) + (x + 4) - (x - 6) - 1

Expanding:

(x + 4)(x - 6) + [ (x + 4) - (x - 6) ] - 1

Compute the bracket: (x + 4 - x + 6) = 10

So numerator becomes: (x + 4)(x - 6) + 10 - 1 = (x + 4)(x - 6) + 9

Therefore, the second term is:

$$
\frac{(x + 4)(x - 6) + 9}{9(x + 4)(x - 6)} = \frac{(x + 4)(x - 6)}{9(x + 4)(x - 6)} + \frac{9}{9(x + 4)(x - 6)} = \frac{1}{9} + \frac{1}{(x + 4)(x - 6)}
$$

Again, let me check with x = 0:

Original second term: (0 + 3)(0 - 5)/[9(0 + 4)(0 - 6)] = (3)(-5)/[9*4*(-6)] = (-15)/(-216) = 5/72.

Simplified expression: 1/9 + 1/[(0 + 4)(0 - 6)] = 1/9 + 1/(4*(-6)) = 1/9 - 1/24 = (8 - 3)/72 = 5/72. Correct! Nice.

Third term: $- \frac{2(x + 5)(x - 7)}{13(x + 6)(x - 8)}$

Let me try the same approach. Let me write numerator as [(x + 6) - 1][(x - 8) + 1]. Let me check:

(x + 5) = (x + 6) - 1, and (x - 7) = (x - 8) + 1. So:

Numerator: [(x + 6) - 1][(x - 8) + 1] = (x + 6)(x - 8) + (x + 6) - (x - 8) - 1

Expanding:

(x + 6)(x - 8) + [ (x + 6) - (x - 8) ] - 1

Compute the bracket: (x + 6 - x + 8) = 14

So numerator becomes: (x + 6)(x - 8) + 14 - 1 = (x + 6)(x - 8) + 13

Therefore, the third term is:

$$
- \frac{2[(x + 6)(x - 8) + 13]}{13(x + 6)(x - 8)} = - \left[ \frac{2(x + 6)(x - 8)}{13(x + 6)(x - 8)} + \frac{2*13}{13(x + 6)(x - 8)} \right] = - \left[ \frac{2}{13} + \frac{2}{(x + 6)(x - 8)} \right] = -\frac{2}{13} - \frac{2}{(x + 6)(x - 8)}
$$

Let me check with x = 0 again:

Original third term: -2*(0 + 5)(0 - 7)/[13*(0 + 6)(0 - 8)] = -2*(5)(-7)/[13*6*(-8)] = -2*(-35)/[13*(-48)] = 70 / (-624) = -35/312.

Simplified expression: -2/13 - 2/[(0 + 6)(0 - 8)] = -2/13 - 2/(6*(-8)) = -2/13 - 2/(-48) = -2/13 + 1/24. Let me compute that:

Convert to common denominator, which is 312. -2/13 = -48/312, 1/24 = 13/312. So total is (-48 + 13)/312 = -35/312. Correct! Perfect.

So now, substituting all three terms back into the original equation:

First term: 1/5 + 1/[(x + 2)(x - 4)]

Second term: 1/9 + 1/[(x + 4)(x - 6)]

Third term: -2/13 - 2/[(x + 6)(x - 8)]

So the entire equation becomes:

$$
\left( \frac{1}{5} + \frac{1}{(x + 2)(x - 4)} \right) + \left( \frac{1}{9} + \frac{1}{(x + 4)(x - 6)} \right) - \left( \frac{2}{13} + \frac{2}{(x + 6)(x - 8)} \right) = \frac{92}{585}
$$

Let me simplify the constants first. Let me compute 1/5 + 1/9 - 2/13.

Compute 1/5 + 1/9: common denominator is 45. 9/45 + 5/45 = 14/45.

Then subtract 2/13: 14/45 - 2/13. Common denominator is 585 (since 45*13=585). So:

14/45 = (14*13)/585 = 182/585

2/13 = (2*45)/585 = 90/585

So 182/585 - 90/585 = 92/585.

Wait! That's exactly the right-hand side of the equation. So the constants add up to 92/585. Therefore, the equation simplifies to:

$$
\frac{92}{585} + \frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = \frac{92}{585}
$$

Subtracting 92/585 from both sides:

$$
\frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = 0
$$

So now, the equation reduces to:

$$
\frac{1}{(x + 2)(x - 4)} + \frac{1}{(x + 4)(x - 6)} - \frac{2}{(x + 6)(x - 8)} = 0
$$

This seems more manageable. Let me denote each term as follows:

Let me call A = 1/[(x + 2)(x - 4)], B = 1/[(x + 4)(x - 6)], C = 2/[(x + 6)(x - 8)]. So equation is A + B - C = 0.

But maybe I can combine these terms. Let me try to combine them step by step.

First, let me combine A and B. Let me find a common denominator for A and B. The denominators are (x + 2)(x - 4) and (x + 4)(x - 6). The common denominator would be (x + 2)(x - 4)(x + 4)(x - 6). But that might be complicated. Alternatively, maybe there's a pattern here.

Alternatively, notice that each denominator is similar to the previous one, shifted by 2 in x. Let me check:

First denominator: (x + 2)(x - 4). Let me write it as (x + 2)(x - 4) = (x + 2)(x - 4)

Second denominator: (x + 4)(x - 6) = (x + 4)(x - 6)

Third denominator: (x + 6)(x - 8) = (x + 6)(x - 8)

So each denominator is (x + 2 + 2k)(x - 4 - 2k) for k=0,1,2. So similar to before.

Alternatively, maybe I can write each term as a difference of fractions. Let me think. For example, similar to how we did before with the first term.

Let me try to express each term as a difference. Let me take the first term: 1/[(x + 2)(x - 4)]. Let me try to write this as something like [A/(x + 2) - B/(x - 4)].

Let me suppose that:

1/[(x + 2)(x - 4)] = A/(x + 2) + B/(x - 4). Wait, but if I want to write it as a difference, maybe:

Let me think of partial fractions. Let me recall that:

1/[(x + a)(x + b)] = [1/(a - b)] [1/(x + b) - 1/(x + a)]

Wait, let me check:

Let me suppose that 1/[(x + a)(x + b)] = C/(x + a) + D/(x + b). Then:

1 = C(x + b) + D(x + a). Let me set x = -a: 1 = C(0) + D(-a + a) = D(0)? Wait, no. Wait, if x = -a, then:

1 = C( -a + b ) + D(0) => C = 1/(b - a). Similarly, x = -b: 1 = C(0) + D(-b + a) => D = 1/(a - b). Therefore:

1/[(x + a)(x + b)] = [1/(b - a)] [1/(x + a) - 1/(x + b)]

So in our case, for the first term, denominator is (x + 2)(x - 4). Let me write it as (x - (-2))(x - 4). So a = -2, b = 4. Then:

1/[(x + 2)(x - 4)] = [1/(4 - (-2))] [1/(x + 2) - 1/(x - 4)] = [1/6][1/(x + 2) - 1/(x - 4)]

Similarly for the second term: 1/[(x + 4)(x - 6)] = [1/( -6 - 4 )][1/(x + 4) - 1/(x - 6)]? Wait, let me check:

Wait, using the formula above, with a = -4, b = 6. Then:

1/[(x + 4)(x - 6)] = [1/(6 - (-4))][1/(x + 4) - 1/(x - 6)] = [1/10][1/(x + 4) - 1/(x - 6)]

Similarly, third term: 2/[(x + 6)(x - 8)] = 2 * [1/( -8 - 6 )][1/(x + 6) - 1/(x - 8)] = 2 * [1/(-14)][1/(x + 6) - 1/(x - 8)] = -2/14 [1/(x + 6) - 1/(x - 8)] = -1/7 [1/(x + 6) - 1/(x - 8)]

So substituting back into the equation:

First term: [1/6][1/(x + 2) - 1/(x - 4)]

Second term: [1/10][1/(x + 4) - 1/(x - 6)]

Third term: - [1/7][1/(x + 6) - 1/(x - 8)]

So the equation becomes:

[1/6][1/(x + 2) - 1/(x - 4)] + [1/10][1/(x + 4) - 1/(x - 6)] - [1/7][1/(x + 6) - 1/(x - 8)] = 0

Let me write this out:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) - (1/7)(1/(x + 6) - 1/(x - 8)) = 0

This seems complicated, but maybe there's a telescoping pattern here. Let me expand each term:

First term: (1/6)(1/(x + 2)) - (1/6)(1/(x - 4))

Second term: (1/10)(1/(x + 4)) - (1/10)(1/(x - 6))

Third term: - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8))

So combining all terms:

[ (1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) ] + [ (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) ] + [ - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8)) ] = 0

Now, let me rearrange the terms by their denominators:

Terms with 1/(x + 2): (1/6)(1/(x + 2))

Terms with 1/(x + 4): (1/10)(1/(x + 4))

Terms with 1/(x - 4): - (1/6)(1/(x - 4))

Terms with 1/(x - 6): - (1/10)(1/(x - 6))

Terms with 1/(x + 6): - (1/7)(1/(x + 6))

Terms with 1/(x - 8): + (1/7)(1/(x - 8))

So, maybe there's a pattern here. Let me check if there's a telescoping effect. Let me think about the denominators:

Looking at denominators: x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. It seems like each term is related to the previous one by adding or subtracting 2. Maybe if I can pair terms?

Alternatively, maybe I can combine terms with similar denominators. Let me see.

Alternatively, maybe I can consider that the equation is a combination of terms that can be grouped. Let me think.

Alternatively, maybe I can write all terms with denominators as follows:

Let me note that denominators are x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. Let me think of them as x + 2, x + 4, x - 4, x - 6, x + 6, x - 8. Maybe if I consider variables shifted by some amount?

Alternatively, maybe I can consider that the equation is symmetric in some way? Not sure. Alternatively, maybe I can combine terms step by step.

Let me try to combine the first two terms:

First, let me consider the terms with 1/(x + 2) and 1/(x + 4). But they are separate. Maybe not helpful.

Alternatively, let me think of the entire expression as:

Let me denote:

Let me write all terms:

Term1: (1/6)(1/(x + 2))

Term2: (1/10)(1/(x + 4))

Term3: - (1/6)(1/(x - 4))

Term4: - (1/10)(1/(x - 6))

Term5: - (1/7)(1/(x + 6))

Term6: + (1/7)(1/(x - 8))

So maybe I can group Term1 and Term3, Term2 and Term4, Term5 and Term6?

Let me try:

Group Term1 and Term3: (1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) = (1/6)[1/(x + 2) - 1/(x - 4)]

Similarly, Group Term2 and Term4: (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) = (1/10)[1/(x + 4) - 1/(x - 6)]

Group Term5 and Term6: - (1/7)(1/(x + 6)) + (1/7)(1/(x - 8)) = (1/7)[ -1/(x + 6) + 1/(x - 8) ] = (1/7)[1/(x - 8) - 1/(x + 6)]

Wait, but this brings us back to the original expression. So that doesn't help. Maybe another grouping?

Alternatively, maybe consider that denominators are spaced by 2, so maybe if I consider variables shifted by 2?

Alternatively, maybe I can consider substituting y = x - something. Let me think. Let me check if there's a substitution that can make denominators symmetric.

Alternatively, maybe I can combine all terms over a common denominator. But that might be very complicated. Let me think.

Alternatively, maybe I can notice that denominators are (x + 2), (x + 4), (x - 4), (x - 6), (x + 6), (x - 8). Let me note that (x - 4) = (x + 2) - 6, but not sure.

Alternatively, maybe I can consider that denominators are related to each other. For example, (x + 2) and (x - 4) differ by 6, (x + 4) and (x - 6) differ by 10, etc. Not sure.

Alternatively, maybe I can consider that the entire equation is a combination of terms that can be telescoped when considering more terms. Wait, but there are only three terms. Let me think again.

Alternatively, maybe I can consider that the equation is similar to a telescoping series. Let me think of the terms as differences. For example, if I have terms like 1/(x + 2) - 1/(x - 4), etc., but with coefficients.

Alternatively, maybe I can write the entire equation as:

Let me think of the entire expression as:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Wait, that's the same as before. So maybe I can write this as:

Let me denote:

Let me think of each bracket as a difference. Let me write:

Let me consider the first bracket: 1/(x + 2) - 1/(x - 4)

Second bracket: 1/(x + 4) - 1/(x - 6)

Third bracket: 1/(x - 8) - 1/(x + 6)

But with coefficients 1/6, 1/10, 1/7 respectively.

But how can I combine these? Maybe if I can find a common pattern or relation between these brackets?

Alternatively, maybe I can consider that each bracket is similar to the previous one, shifted by 2. Let me check:

First bracket: 1/(x + 2) - 1/(x - 4)

Second bracket: 1/(x + 4) - 1/(x - 6)

Third bracket: 1/(x - 8) - 1/(x + 6)

If I look at the first bracket, if I shift x by 2, I get the second bracket. Similarly, shifting x by 2 again, I get something else. But not sure.

Alternatively, maybe I can consider that the entire expression is a combination of terms that can be telescoped when considering more terms. But since there are only three terms, maybe not.

Alternatively, maybe I can consider that the equation is equal to zero, so:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Let me write all terms:

(1/6)(1/(x + 2)) - (1/6)(1/(x - 4)) + (1/10)(1/(x + 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) - (1/7)(1/(x + 6)) = 0

Now, let me rearrange the terms:

[ (1/6)(1/(x + 2)) + (1/10)(1/(x + 4)) - (1/7)(1/(x + 6)) ] + [ - (1/6)(1/(x - 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) ] = 0

Hmm, not sure. Maybe I can consider that the terms with positive signs and negative signs are separate. Let me think.

Alternatively, maybe I can consider that denominators are x + 2, x + 4, x + 6, x - 4, x - 6, x - 8. Let me note that x - 4 = (x + 2) - 6, x - 6 = (x + 4) - 10, x - 8 = (x + 6) - 14. Not sure.

Alternatively, maybe I can make a substitution. Let me set y = x - something. Let me think. For example, if I set y = x - 3, maybe? Let me try.

But maybe not. Alternatively, maybe set t = x - something to make denominators symmetric. Let me think.

Alternatively, maybe I can consider that denominators are (x + 2), (x + 4), (x + 6), and their counterparts (x - 4), (x - 6), (x - 8). Let me note that (x - 4) = (x + 2) - 6, (x - 6) = (x + 4) - 10, (x - 8) = (x + 6) - 14. Not sure.

Alternatively, maybe I can consider that denominators are in arithmetic progression? For example, x + 2, x + 4, x + 6, etc. But the other denominators are x - 4, x - 6, x - 8. So maybe if I consider two sequences: one increasing by 2, and one decreasing by 2.

Alternatively, maybe I can think of the entire equation as a combination of terms that can be expressed in terms of 1/(x + 2), 1/(x + 4), 1/(x + 6), etc. Let me try to write all terms in terms of these.

Let me note that:

- (1/6)(1/(x - 4)) = - (1/6)(1/(x + 2 - 6)) = - (1/6)(1/(x + 2 - 6))

Similarly, - (1/10)(1/(x - 6)) = - (1/10)(1/(x + 4 - 10))

And + (1/7)(1/(x - 8)) = + (1/7)(1/(x + 6 - 14))

But this seems too vague. Maybe not helpful.

Alternatively, maybe I can consider that the equation is a linear combination of terms like 1/(x + 2), 1/(x + 4), 1/(x + 6), 1/(x - 4), 1/(x - 6), 1/(x - 8). Let me denote:

Let me write the equation as:

(1/6)(1/(x + 2)) + (1/10)(1/(x + 4)) - (1/7)(1/(x + 6)) - (1/6)(1/(x - 4)) - (1/10)(1/(x - 6)) + (1/7)(1/(x - 8)) = 0

Let me group terms with similar denominators:

For 1/(x + 2): (1/6)

For 1/(x + 4): (1/10)

For 1/(x + 6): - (1/7)

For 1/(x - 4): - (1/6)

For 1/(x - 6): - (1/10)

For 1/(x - 8): + (1/7)

So maybe I can write this as:

(1/6)(1/(x + 2) - 1/(x - 4)) + (1/10)(1/(x + 4) - 1/(x - 6)) + (1/7)(1/(x - 8) - 1/(x + 6)) = 0

Which is the same as before. So I need another approach.

Alternatively, maybe I can consider that the equation is equal to zero, so:

Let me denote:

Let me consider that denominators are (x + 2), (x + 4), (x + 6), (x - 4), (x - 6), (x - 8). Let me think of them as pairs:

(x + 2) and (x - 4), (x + 4) and (x - 6), (x + 6) and (x - 8). Each pair differs by 6, 10, 14? Wait, (x + 2) and (x - 4): difference is 6. Similarly, (x + 4) and (x - 6): difference is 10. (x + 6) and (x - 8): difference is 14. So differences are 6, 10, 14. Which are increasing by 4 each time. Interesting. So maybe there's a pattern here.

But how does that help? Maybe not directly.

Alternatively, maybe I can consider that each pair is related to the previous one. For example, the first pair (x + 2) and (x - 4) can be thought of as (x + 2) and (x - 4), then the next pair is (x + 4) and (x - 6), which is similar but shifted by 2. Maybe if I consider that the entire equation is a telescoping sum?

Alternatively, maybe I can consider that the equation is a combination of terms that can be expressed as differences of fractions with denominators that telescope when combined. For example, if I can write the entire expression as a single fraction, maybe it simplifies.

But combining all terms would be very tedious. Let me try.

First, let me note that denominators are:

For the first term: (x + 2)(x - 4)

Second term: (x + 4)(x - 6)

Third term: (x + 6)(x - 8)

So the common denominator would be the product of all three denominators: (x + 2)(x - 4)(x + 4)(x - 6)(x + 6)(x - 8). That's a huge denominator. But maybe manageable?

Alternatively, maybe I can combine two terms at a time. Let me try combining the first two terms first.

Let me take the first two terms:

Term1 + Term2 = [1/(x + 2)(x - 4)] + [1/(x + 4)(x - 6)]

Let me find a common denominator for these two terms: (x + 2)(x - 4)(x + 4)(x - 6). Let me compute:

Term1: [ (x + 4)(x - 6) ] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] + [ (x + 2)(x - 4) ] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ]

So numerator: (x + 4)(x - 6) + (x + 2)(x - 4)

Let me compute that:

First part: (x + 4)(x - 6) = x² - 6x + 4x - 24 = x² - 2x - 24

Second part: (x + 2)(x - 4) = x² - 4x + 2x - 8 = x² - 2x - 8

Adding them together: (x² - 2x - 24) + (x² - 2x - 8) = 2x² - 4x - 32

Therefore, Term1 + Term2 = [2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ]

Simplify numerator: 2x² - 4x - 32 = 2(x² - 2x - 16). Not sure if factors. Let me check discriminant: 4 + 64 = 68. Not a perfect square. So maybe leave as is.

Now, adding Term3: - [2 / (x + 6)(x - 8)]

So the entire equation is:

[2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] - [2 / (x + 6)(x - 8)] = 0

Let me write this as:

[2x² - 4x - 32] / [ (x + 2)(x - 4)(x + 4)(x - 6) ) ] = [2 / (x + 6)(x - 8)]

Multiply both sides by (x + 2)(x - 4)(x + 4)(x - 6)(x + 6)(x - 8) to eliminate denominators:

Left side: [2x² - 4x - 32] * (x + 6)(x - 8)

Right side: 2 * (x + 2)(x - 4)(x + 4)(x - 6)

So equation becomes:

(2x² - 4x - 32)(x + 6)(x - 8) = 2(x + 2)(x - 4)(x + 4)(x - 6)

Let me simplify both sides.

First, factor out 2 from the left side:

Left side: 2(x² - 2x - 16)(x + 6)(x - 8)

Right side: 2(x + 2)(x - 4)(x + 4)(x - 6)

We can divide both sides by 2:

(x² - 2x - 16)(x + 6)(x - 8) = (x + 2)(x - 4)(x + 4)(x - 6)

Now, let me compute both sides.

First, let me compute the right side: (x + 2)(x - 4)(x + 4)(x - 6)

Note that (x + 2)(x - 4) = x² - 2x - 8 and (x + 4)(x - 6) = x² - 2x - 24. So the right side is (x² - 2x - 8)(x² - 2x - 24). Let me denote y = x² - 2x. Then right side becomes (y - 8)(y - 24) = y² - 32y + 192.

Similarly, left side: (x² - 2x - 16)(x + 6)(x - 8). Let me compute (x + 6)(x - 8) = x² - 2x - 48. So left side is (x² - 2x - 16)(x² - 2x - 48). Let me denote y = x² - 2x again. Then left side becomes (y - 16)(y - 48) = y² - 64y + 768.

So now, equation becomes:

Left side: y² - 64y + 768 = Right side: y² - 32y + 192

Subtracting right side from both sides:

(y² - 64y + 768) - (y² - 32y + 192) = 0 => -32y + 576 = 0 => -32y + 576 = 0 => 32y = 576 => y = 576 / 32 = 18.

So y = 18. Recall that y = x² - 2x. Therefore:

x² - 2x = 18 => x² - 2x - 18 = 0.

Solving this quadratic equation:

x = [2 ± sqrt(4 + 72)] / 2 = [2 ± sqrt(76)] / 2 = [2 ± 2*sqrt(19)] / 2 = 1 ± sqrt(19).

So the solutions are x = 1 + sqrt(19) and x = 1 - sqrt(19).

But wait, I need to check if these solutions are valid, i.e., they don't make any denominators zero in the original equation.

Original equation denominators are:

First term: 5(x + 2)(x - 4)

Second term: 9(x + 4)(x - 6)

Third term: 13(x + 6)(x - 8)

So we need to ensure that x ≠ -2, 4, -4, 6, -6, 8.

Let me check for x = 1 + sqrt(19) and x = 1 - sqrt(19).

First, approximate sqrt(19) ≈ 4.3589.

So x ≈ 1 + 4.3589 ≈ 5.3589, and x ≈ 1 - 4.3589 ≈ -3.3589.

Check if these values are equal to -2, 4, -4, 6, -6, 8.

For x ≈ 5.3589: not equal to any of those. For x ≈ -3.3589: not equal to any of those. So both solutions are valid.

Therefore, the real roots are x = 1 + sqrt(19) and x = 1 - sqrt(19).

But let me verify this solution by plugging back into the original equation to ensure no mistakes were made during the algebra.

Let me check with x = 1 + sqrt(19). Let me compute the left-hand side (LHS) and see if it equals 92/585.

But this might be tedious. Alternatively, since we followed the algebra correctly, and the steps were reversible (except for multiplying both sides by denominators, which we need to check for extraneous solutions, but we already checked that the solutions don't make denominators zero), so likely correct.

But let me check with x = 1 + sqrt(19). Let me compute y = x² - 2x = 18. So that was the key step. Since we derived that from the equation, and the rest of the steps were algebraic manipulations, it should be correct.

Therefore, the real roots are 1 ± sqrt(19).

But let me check if there are any other solutions. Wait, when we divided both sides by 2, and then simplified, we arrived at y = 18, leading to quadratic equation. So only two solutions. Therefore, the real roots are 1 + sqrt(19) and 1 - sqrt(19).

But let me check if there was any step where we might have lost solutions or introduced extraneous ones. For example, when we multiplied both sides by denominators, we assumed that denominators are not zero, which we checked. Also, when we divided by 2, that's fine. So I think these are the only solutions.

Therefore, the real roots are 1 + sqrt(19) and 1 - sqrt(19).

But let me check if the original equation simplifies correctly. Let me take x = 1 + sqrt(19). Let me compute the left-hand side.

But maybe it's too time-consuming. Alternatively, since we followed the steps correctly, and the algebra led us to a quadratic equation with two solutions, and they don't make denominators zero, they should be valid.

Therefore, the answer is x = 1 ± sqrt(19). So in boxed form, they should be written as \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

But the problem says "Enter the real roots, separated by commas." So likely need to write both in one box, separated by commas.

**Final Answer**
The real roots are \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

Wait, but the instructions say "Enter the real roots, separated by commas." So maybe inside a single box? Let me check the problem statement again.

"Enter the real roots, separated by commas."

So likely, they want both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system expects multiple boxed answers? Or maybe a single box with both. Let me check standard practice. Usually, for multiple answers, they are separated by commas inside a single box. For example, \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} might be two separate boxes, but if the system allows only one box, then inside one box. Since the original problem says "Enter the real roots, separated by commas," I think they expect both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system requires a single box. Let me check the initial problem statement. The user says "put your final answer within \boxed{}". So maybe they expect multiple boxed answers? But the standard practice here is to have one box. Maybe the answer is two numbers, so written as \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.

But since the user might expect comma-separated inside one box, I'll check similar problems. For example, if there are multiple solutions, they are often listed in one box separated by commas. So I think:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But maybe the system requires a single box. Let me check the original problem's example. The original equation is given, and the user says "Enter the real roots, separated by commas." So likely, they want both answers in one box, separated by commas. So:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But actually, in many cases, for multiple answers, they are put in one box. For example, on AoPS or other platforms, multiple answers are separated by commas in a single box. So I think the correct format is:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since the user might have specific instructions, but given the initial problem says "put your final answer within \boxed{}", maybe they expect multiple boxed answers. However, since the user hasn't specified, but in the problem statement says "Enter the real roots, separated by commas," I think safest is to put both in one box separated by commas.

**Final Answer**
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

Wait, but actually, looking at the problem again, the original equation is equal to 92/585. When I simplified, I found that the constants added up to 92/585, so the equation reduced to the sum of the fractions equals zero. Then through substitutions, I arrived at the quadratic equation. So the solutions are correct. Therefore, the real roots are 1 ± sqrt(19). So I think the answer is those two values. So I'll go with:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But if the system requires a single box, maybe:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since the user might have intended multiple boxed answers, but the instruction says "put your final answer within \boxed{}", which is singular. Maybe they expect both answers in one box. For example:

\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

Yes, that's possible. Let me check standard practice. In many cases, for multiple answers, they are put in one box separated by commas. For example, on AoPS or math competitions, they might write like that. So I think that's the way to go.

**Final Answer**
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

Wait, but the system might not accept multiple boxes. Since the user hasn't specified, but given the initial problem says "Enter the real roots, separated by commas," I think the correct format is:

\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}

But since I can't be sure, but given the original instruction says "put your final answer within \boxed{}", maybe they expect both answers in one box. So:

\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

Yes, that's likely. So I'll go with that.

**Final Answer**
\boxed{1 + \sqrt{19}, 1 - \sqrt{19}}

To solve the equation:

$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} + \frac{(x+3)(x-5)}{9(x+4)(x-6)} - \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = \frac{92}{585}
$$

we begin by simplifying each term using a clever algebraic technique. Each term has a numerator and denominator that are products of linear expressions, and we observe that the numerators and denominators are related by a consistent pattern.

---

### Step 1: Simplify Each Term

We use the identity:

$$
\frac{(x + a)(x - b)}{(x + c)(x - d)} = \frac{1}{(c - d)} \left( \frac{1}{x + c} - \frac{1}{x - d} \right)
$$

This allows us to rewrite each term as a difference of two simpler fractions.

#### First Term:
$$
\frac{(x+1)(x-3)}{5(x+2)(x-4)} = \frac{1}{5} + \frac{1}{(x+2)(x-4)}
$$

#### Second Term:
$$
\frac{(x+3)(x-5)}{9(x+4)(x-6)} = \frac{1}{9} + \frac{1}{(x+4)(x-6)}
$$

#### Third Term:
$$
- \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = -\frac{2}{13} - \frac{2}{(x+6)(x-8)}
$$

---

### Step 2: Combine All Terms

Substituting back into the original equation:

$$
\left( \frac{1}{5} + \frac{1}{(x+2)(x-4)} \right) + \left( \frac{1}{9} + \frac{1}{(x+4)(x-6)} \right) - \left( \frac{2}{13} + \frac{2}{(x+6)(x-8)} \right) = \frac{92}{585}
$$

Now, combine the constants:

$$
\frac{1}{5} + \frac{1}{9} - \frac{2}{13} = \frac{92}{585}
$$

This confirms that the constants on the left-hand side match the right-hand side. So the equation reduces to:

$$
\frac{1}{(x+2)(x-4)} + \frac{1}{(x+4)(x-6)} - \frac{2}{(x+6)(x-8)} = 0
$$

---

### Step 3: Use Partial Fractions

We apply the identity again to each term:

- $\frac{1}{(x+2)(x-4)} = \frac{1}{6} \left( \frac{1}{x+2} - \frac{1}{x-4} \right)$
- $\frac{1}{(x+4)(x-6)} = \frac{1}{10} \left( \frac{1}{x+4} - \frac{1}{x-6} \right)$
- $\frac{2}{(x+6)(x-8)} = \frac{2}{14} \left( \frac{1}{x+6} - \frac{1}{x-8} \right) = \frac{1}{7} \left( \frac{1}{x+6} - \frac{1}{x-8} \right)$

Substituting back:

$$
\frac{1}{6} \left( \frac{1}{x+2} - \frac{1}{x-4} \right) + \frac{1}{10} \left( \frac{1}{x+4} - \frac{1}{x-6} \right) - \frac{1}{7} \left( \frac{1}{x+6} - \frac{1}{x-8} \right) = 0
$$

---

### Step 4: Combine All Terms

Grouping all terms:

$$
\frac{1}{6} \cdot \frac{1}{x+2} - \frac{1}{6} \cdot \frac{1}{x-4} + \frac{1}{10} \cdot \frac{1}{x+4} - \frac{1}{10} \cdot \frac{1}{x-6} - \frac{1}{7} \cdot \frac{1}{x+6} + \frac{1}{7} \cdot \frac{1}{x-8} = 0
$$

Now, we combine all terms over a common denominator. However, instead of expanding, we observe that the entire expression simplifies to a quadratic equation in terms of $x^2 - 2x$.

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

$$
(y - 16)(y - 48) = (y - 8)(y - 24)
$$

Expanding both sides:

- Left: $y^2 - 64y + 768$
- Right: $y^2 - 32y + 192$

Subtracting:

$$
(y^2 - 64y + 768) - (y^2 - 32y + 192) = -32y + 576 = 0
\Rightarrow y = 18
$$

Recall $y = x^2 - 2x$, so:

$$
x^2 - 2x = 18 \Rightarrow x^2 - 2x - 18 = 0
$$

Solving this quadratic:

$$
x = \frac{2 \pm \sqrt{4 + 72}}{2} = \frac{2 \pm \sqrt{76}}{2} = 1 \pm \sqrt{19}
$$

---

### Step 5: Final Answer

We check that these values do not make any denominator zero in the original equation. Since the denominators are:

- $x + 2$, $x - 4$, $x + 4$, $x - 6$, $x + 6$, $x - 8$

and $1 \pm \sqrt{19} \approx 5.3589$ and $-3.3589$, none of these values are equal to any of the excluded values, so both are valid.

---

### ✅ Final Answer:

$$
\boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}
$$

Solution:
Multiplying out each numerator and denominator, we get
\[\frac{x^2 - 2x - 3}{5(x^2 - 2x - 8)} + \frac{x^2 - 2x - 15}{9(x^2 - 2x - 24)} - \frac{2(x^2 - 2x - 35)}{13(x^2 - 2x - 48)} = \frac{92}{585}.\]We can write this as
\[\frac{(x^2 - 2x - 8) + 5}{5(x^2 - 2x - 8)} + \frac{(x^2 - 2x - 24) + 9}{9(x^2 - 2x - 24)} - \frac{2((x^2 - 2x - 48) + 13)}{13(x^2 - 2x - 48)} = \frac{92}{585}.\]Hence,
\[\frac{1}{5} + \frac{1}{x^2 - 2x - 8} + \frac{1}{9} + \frac{1}{x^2 - 2x - 24} - \frac{2}{13} - \frac{2}{x^2 - 2x - 48} = \frac{92}{585}.\]This simplifies to
\[\frac{1}{x^2 - 2x - 8} + \frac{1}{x^2 - 2x - 24} - \frac{2}{x^2 - 2x - 48} = 0.\]Let $y = x^2 - 2x - 48.$  Then
\[\frac{1}{y + 40} + \frac{1}{y + 24} - \frac{2}{y} = 0.\]Multiplying everything by $y(y + 24)(y + 40),$ we get
\[y(y + 24) + y(y + 40) - 2(y + 24)(y + 40) = 0.\]This simplifies to $64y + 1920 = 0,$ so $y = -30.$  Then $x^2 - 2x - 48 = -30,$ or $x^2 - 2x - 18 = 0.$  By the quadratic formula, $x = \boxed{1 \pm \sqrt{19}}.$  (Since the denominators are nonzero for these values, we know that they are not extraneous.)