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 expanding didn't help. Maybe I can write each term as something over (x + something)(x - something). Let me think.

Alternatively, maybe notice that each term has a similar structure. Let me check:

First term numerator: (x + 1)(x - 3) = x² - 2x - 3. Denominator: 5(x + 2)(x - 4) = 5(x² - 2x - 8). So if I write numerator as (x² - 2x - 8) + 5. Because x² - 2x - 8 + 5 = x² - 2x - 3. So numerator is denominator/5 + 5? Wait:

Let me check:

Denominator is 5(x² - 2x - 8). So if I take denominator divided by 5, that's x² - 2x - 8. Then adding 5 gives x² - 2x - 8 + 5 = x² - 2x - 3, which is the numerator. So numerator = (denominator / 5) + 5. Therefore, the first term can be written as:

[ (denominator / 5) + 5 ] / denominator = (denominator / 5)/denominator + 5 / denominator = 1/5 + 5 / denominator.

Wait, that seems useful! Let me verify:

Let me denote D1 = 5(x + 2)(x - 4). Then numerator is (x + 1)(x - 3) = D1 / 5 + 5? Let me check:

D1 / 5 = (x + 2)(x - 4) = x² - 2x - 8. Then adding 5 gives x² - 2x - 8 + 5 = x² - 2x - 3, which is the numerator. Yes! So the first term is [D1 / 5 + 5]/D1 = 1/5 + 5/D1.

Similarly, maybe the same approach can be applied to the other terms?

Let me check the second term: numerator is (x + 3)(x - 5) = x² - 2x -15. Denominator is 9(x + 4)(x - 6) = 9(x² - 2x - 24). Let me check if numerator can be expressed as denominator / 9 + something.

Denominator / 9 = (x + 4)(x - 6) = x² - 2x - 24. Then adding something to get numerator x² - 2x -15. So x² - 2x -24 + something = x² - 2x -15. Therefore, something is 9. So numerator = denominator / 9 + 9. Therefore, the second term is [denominator / 9 + 9]/denominator = 1/9 + 9 / denominator.

Similarly for the third term: numerator is 2(x + 5)(x - 7) = 2(x² - 2x - 35). Denominator is 13(x + 6)(x - 8) = 13(x² - 2x - 48). Let me check if numerator can be expressed as denominator / 13 + something.

Denominator / 13 = (x + 6)(x - 8) = x² - 2x - 48. Then adding something to get numerator 2(x² - 2x - 35) = 2x² - 4x -70. Wait, but denominator / 13 is x² - 2x - 48. So if I multiply denominator / 13 by 2, I get 2x² - 4x - 96. Then adding something to get 2x² -4x -70. So 2x² -4x -96 + something = 2x² -4x -70. Therefore, something is 26. Therefore, numerator = 2*(denominator / 13) + 26. Therefore, the third term is [2*(denominator / 13) + 26]/denominator = 2/13 + 26 / denominator.

Wait, but the third term is subtracted. Let me check:

Original third term is - [2(x + 5)(x - 7)] / [13(x + 6)(x - 8)] = - [2*(denominator / 13) + 26]/denominator = - [2/13 + 26 / denominator]. So that would be -2/13 - 26 / denominator.

So putting all together, the original equation becomes:

First term: 1/5 + 5 / D1

Second term: 1/9 + 9 / D2

Third term: -2/13 - 26 / D3

So adding them all up:

(1/5 + 1/9 - 2/13) + (5/D1 + 9/D2 - 26/D3) = 92/585.

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

To compute that, find common denominators. Let me compute:

First, 1/5 + 1/9 = (9 + 5)/45 = 14/45.

Then subtract 2/13: 14/45 - 2/13. Let me compute that:

Find common denominator of 45 and 13, which is 585.

Convert 14/45 to (14 * 13)/585 = 182/585.

Convert 2/13 to (2 * 45)/585 = 90/585.

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

Wait! So the constants add up to 92/585. Therefore, the equation becomes:

92/585 + (5/D1 + 9/D2 - 26/D3) = 92/585.

Subtracting 92/585 from both sides:

5/D1 + 9/D2 - 26/D3 = 0.

So now, the equation simplifies to:

5/D1 + 9/D2 - 26/D3 = 0.

Where D1 = 5(x + 2)(x - 4), D2 = 9(x + 4)(x - 6), D3 = 13(x + 6)(x - 8).

Wait, but let me check:

Earlier, we had:

First term: 1/5 + 5/D1

Second term: 1/9 + 9/D2

Third term: -2/13 - 26/D3

So adding them up:

(1/5 + 1/9 - 2/13) + (5/D1 + 9/D2 - 26/D3) = 92/585.

But we found that (1/5 + 1/9 - 2/13) = 92/585, so the equation reduces to:

92/585 + (5/D1 + 9/D2 - 26/D3) = 92/585 => 5/D1 + 9/D2 - 26/D3 = 0.

So now, the problem reduces to solving:

5/D1 + 9/D2 - 26/D3 = 0.

But D1, D2, D3 are denominators from original terms. Let me write them again:

D1 = 5(x + 2)(x - 4)

D2 = 9(x + 4)(x - 6)

D3 = 13(x + 6)(x - 8)

So substituting:

5 / [5(x + 2)(x - 4)] + 9 / [9(x + 4)(x - 6)] - 26 / [13(x + 6)(x - 8)] = 0.

Simplify each term:

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

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

Third term: -26 / [13(x + 6)(x - 8)] = -2 / [(x + 6)(x - 8)]

So the equation simplifies to:

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

That's a big simplification! So now, the equation is:

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

This seems more manageable. Let me write this equation again:

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

Now, perhaps I can combine these fractions. To do that, I need a common denominator. However, the denominators are all quadratic terms. Let me note that each denominator is of the form (x + a)(x - b), where a and b are constants. Let me check if there's a pattern or substitution.

Alternatively, maybe I can use partial fractions or some substitution. Let me think. Alternatively, notice that each denominator is similar to the previous one with shifted terms. Let me check:

First denominator: (x + 2)(x - 4) = x² - 2x - 8

Second denominator: (x + 4)(x - 6) = x² - 2x - 24

Third denominator: (x + 6)(x - 8) = x² - 2x - 48

Wait, interesting! Each denominator is x² - 2x - something. Let me denote y = x² - 2x. Then denominators are y - 8, y - 24, y - 48. So maybe substitution?

Let me try that. Let me set y = x² - 2x. Then denominators become:

First: y - 8

Second: y - 24

Third: y - 48

So the equation becomes:

1/(y - 8) + 1/(y - 24) - 2/(y - 48) = 0.

That seems promising! Let me write that:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} - \frac{2}{y - 48} = 0
$$

Now, solving for y. Let me combine these terms. Let me first combine the first two terms:

1/(y - 8) + 1/(y - 24) = [ (y - 24) + (y - 8) ] / [(y - 8)(y - 24)] = [2y - 32]/[(y - 8)(y - 24)].

So the equation becomes:

[2y - 32]/[(y - 8)(y - 24)] - 2/(y - 48) = 0.

Let me write that as:

[2(y - 16)] / [(y - 8)(y - 24)] - 2/(y - 48) = 0.

Factor out the 2:

2[ (y - 16)/[(y - 8)(y - 24)] - 1/(y - 48) ] = 0.

Since 2 ≠ 0, we can divide both sides by 2:

(y - 16)/[(y - 8)(y - 24)] - 1/(y - 48) = 0.

So:

(y - 16)/[(y - 8)(y - 24)] = 1/(y - 48).

Cross-multiplying:

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

Let me expand both sides.

Left side: (y - 16)(y - 48) = y² - 48y -16y + 16*48 = y² - 64y + 768.

Right side: (y - 8)(y - 24) = y² -24y -8y + 192 = y² -32y + 192.

Set left side equal to right side:

y² - 64y + 768 = y² -32y + 192.

Subtract y² from both sides:

-64y + 768 = -32y + 192.

Bring all terms to left side:

-64y + 768 + 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, we need to check if these solutions are valid, i.e., they don't make any denominators zero in the original equation.

Original denominators in the simplified equation were (x + 2)(x - 4), (x + 4)(x - 6), (x + 6)(x - 8). Let me check for x = 1 ± sqrt(19).

First, compute approximate values to check if they are in the domain.

sqrt(19) is approx 4.3589.

So x = 1 + 4.3589 ≈ 5.3589

x = 1 - 4.3589 ≈ -3.3589

Check denominators for x ≈ 5.3589:

First denominator: (x + 2)(x - 4) ≈ (5.3589 + 2)(5.3589 - 4) ≈ (7.3589)(1.3589) ≈ not zero.

Second denominator: (x + 4)(x - 6) ≈ (5.3589 + 4)(5.3589 - 6) ≈ (9.3589)(-0.6411) ≈ not zero.

Third denominator: (x + 6)(x - 8) ≈ (5.3589 + 6)(5.3589 - 8) ≈ (11.3589)(-2.6411) ≈ not zero.

Similarly for x ≈ -3.3589:

First denominator: (x + 2)(x - 4) ≈ (-3.3589 + 2)(-3.3589 - 4) ≈ (-1.3589)(-7.3589) ≈ not zero.

Second denominator: (x + 4)(x - 6) ≈ (-3.3589 + 4)(-3.3589 - 6) ≈ (0.6411)(-9.3589) ≈ not zero.

Third denominator: (x + 6)(x - 8) ≈ (-3.3589 + 6)(-3.3589 - 8) ≈ (2.6411)(-11.3589) ≈ not zero.

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

But let me check if there are any other solutions. Wait, when we did the substitution, we had y = x² - 2x, and solved for y = 18, leading to quadratic equation. So only two solutions. Therefore, these are the only real roots.

But let me check if there was any step where we might have introduced extraneous solutions or lost solutions. For example, when we cross-multiplied, we assumed that denominators are not zero. But we already checked that our solutions don't make denominators zero, so they are valid.

Alternatively, let me check by plugging back into the original equation. Let me take x = 1 + sqrt(19). Let me compute the left-hand side (LHS) and see if it equals 92/585.

But that might be tedious. Alternatively, since we followed algebraic steps correctly, and checked that the solutions don't make denominators zero, they should be valid. However, to be thorough, let me check with approximate values.

First, compute approximate value of sqrt(19) ≈ 4.3589, so x ≈ 5.3589.

Compute each term:

First term: [(x+1)(x-3)] / [5(x+2)(x-4)]

Compute numerator: (5.3589 + 1)(5.3589 - 3) ≈ (6.3589)(2.3589) ≈ approx 6.3589 * 2.3589 ≈ let me compute 6 * 2.3589 = 14.1534, 0.3589 * 2.3589 ≈ approx 0.846, total ≈ 14.1534 + 0.846 ≈ 14.999 ≈ 15.

Denominator: 5*(5.3589 + 2)*(5.3589 - 4) = 5*(7.3589)*(1.3589) ≈ 5*(7.3589*1.3589). Let me compute 7 * 1.3589 ≈ 9.5123, 0.3589*1.3589 ≈ approx 0.487, total ≈ 9.5123 + 0.487 ≈ 10. So denominator ≈ 5*10 = 50. Therefore, first term ≈ 15 / 50 = 0.3.

Second term: [(x+3)(x-5)] / [9(x+4)(x-6)]

Numerator: (5.3589 + 3)(5.3589 - 5) ≈ (8.3589)(0.3589) ≈ approx 8 * 0.3589 = 2.8712 + 0.3589*0.3589 ≈ approx 2.8712 + 0.128 ≈ 3.0.

Denominator: 9*(5.3589 + 4)*(5.3589 - 6) = 9*(9.3589)*(-0.6411) ≈ 9*( -5.999 ) ≈ -53.991. Therefore, second term ≈ 3 / (-53.991) ≈ -0.0556.

Third term: - [2(x+5)(x-7)] / [13(x+6)(x-8)]

Numerator: 2*(5.3589 + 5)*(5.3589 - 7) = 2*(10.3589)*(-1.6411) ≈ 2*(-16.98) ≈ -33.96.

Denominator: 13*(5.3589 + 6)*(5.3589 - 8) = 13*(11.3589)*(-2.6411) ≈ 13*(-29.96) ≈ -389.48.

Third term ≈ - [ -33.96 / -389.48 ] = - [ 0.0872 ] ≈ -0.0872.

So total LHS ≈ 0.3 - 0.0556 - 0.0872 ≈ 0.3 - 0.1428 ≈ 0.1572.

Now, 92/585 ≈ let me compute that. 585 divided by 92: 585 ÷ 92 ≈ 6.3587. So 1/6.3587 ≈ 0.1572. Therefore, 92/585 ≈ 0.1572. Which matches the approximate value of LHS. Therefore, x ≈ 5.3589 is a valid solution.

Similarly, check x ≈ -3.3589.

First term: [(x+1)(x-3)] / [5(x+2)(x-4)]

Numerator: (-3.3589 + 1)(-3.3589 - 3) = (-2.3589)(-6.3589) ≈ 15.

Denominator: 5*(-3.3589 + 2)*(-3.3589 - 4) = 5*(-1.3589)*(-7.3589) ≈ 5*(10) = 50. So first term ≈ 15 / 50 = 0.3.

Second term: [(x+3)(x-5)] / [9(x+4)(x-6)]

Numerator: (-3.3589 + 3)(-3.3589 - 5) = (-0.3589)(-8.3589) ≈ 2.999 ≈ 3.

Denominator: 9*(-3.3589 + 4)*(-3.3589 - 6) = 9*(0.6411)*(-9.3589) ≈ 9*(-5.999) ≈ -53.991. So second term ≈ 3 / (-53.991) ≈ -0.0556.

Third term: - [2(x+5)(x-7)] / [13(x+6)(x-8)]

Numerator: 2*(-3.3589 + 5)*(-3.3589 - 7) = 2*(1.6411)*(-10.3589) ≈ 2*(-16.98) ≈ -33.96.

Denominator: 13*(-3.3589 + 6)*(-3.3589 - 8) = 13*(2.6411)*(-11.3589) ≈ 13*(-29.96) ≈ -389.48.

Third term ≈ - [ -33.96 / -389.48 ] = - [ 0.0872 ] ≈ -0.0872.

So total LHS ≈ 0.3 - 0.0556 - 0.0872 ≈ same as before, 0.1572 ≈ 92/585. Therefore, x ≈ -3.3589 is also a valid solution.

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

But let me check if there are any other solutions. For example, when we did substitution y = x² - 2x, and solved for y = 18, leading to quadratic equation. So only two solutions. Therefore, these are the only real roots.

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

But the problem says "Enter the real roots, separated by commas." So maybe as \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}? Or maybe in a single box? Wait, the original instruction says "Enter the real roots, separated by commas." So likely, they want both roots in one box, separated by commas. Let me check the problem statement again:

"Enter the real roots, separated by commas."

So probably, the answer should be written as \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}? Or maybe inside a single box? But standard practice for multiple answers is to put them in separate boxes or in one box separated by commas. Since the user instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the original problem says "Enter the real roots, separated by commas." So maybe inside a single box, separated by commas. Let me check previous problems. For example, if there are multiple answers, they are often put in one box separated by commas. For example, \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} might be two boxes, but maybe the user wants them in one box. However, since the user says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the instruction is ambiguous. However, given that the original equation is likely to have two real roots, and the standard practice here is to put multiple answers in separate boxes. However, since the user might have intended a single box, but I need to check.

Alternatively, maybe I can check if there are more solutions. Wait, we had y = x² - 2x = 18, leading to x² - 2x - 18 = 0, which has two real roots. So only two real roots. Therefore, the answer is those two. So I think the correct way is to write both in boxed form separated by commas. However, since the user says "put your final answer within \boxed{}", maybe they want both answers in one box. For example:

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

But maybe the system expects multiple boxed answers. However, since the original problem is from an Olympiad or similar, and the standard is to write all real roots separated by commas inside a single box. For example, like \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} or \boxed{1 + \sqrt{19}, 1 - \sqrt{19}}.

But I need to check the exact instruction. The user says: "Enter the real roots, separated by commas." and "put your final answer within \boxed{}". So maybe they want both roots inside a single box, separated by commas. For example:

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

But I need to check if that's acceptable. In some contexts, multiple answers are put in separate boxes, but since the user says "within \boxed{}", maybe they want one box. However, the exact instruction is ambiguous. However, given that the original equation is likely to have two real roots, and the standard practice in many competitions is to list all solutions in a single box separated by commas, I will go with that.

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

Wait, but the user might expect both in one box. Let me check again. The original instruction says: "Enter the real roots, separated by commas." So maybe:

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

But I think the standard on platforms like AoPS or others is to put multiple answers in separate boxes. However, since the user hasn't specified, but given the initial problem statement, I think safest is to put both in one box separated by commas. However, in many cases, multiple boxed answers are acceptable. But since the user says "put your final answer within \boxed{}", maybe they want a single box. Let me check the problem again. The original equation is a single equation, so the answer is two real roots. In many cases, for such problems, the answer is written as two boxed expressions. However, since the user might be using an automated system that expects a single box, but I can't be sure. Given that, I think safest is to follow the instruction: "Enter the real roots, separated by commas." So if I have to write them in one box, separated by commas. So:

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

But I need to check if that's acceptable. In some contexts, commas inside the box are okay. For example, on AoPS, they sometimes do that. Alternatively, if the system is expecting multiple boxed answers, but since the user says "put your final answer within \boxed{}", maybe they want one box. Given that, I will go with:

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

But since the user might have intended a single box, but I can't be sure. However, given that the original instruction says "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will write both roots inside a single box separated by commas.

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

Wait, but that's two boxes. Maybe the user wants them in one box. Let me check the exact instruction again: "Enter the real roots, separated by commas." So if I have to enter them separated by commas, then in the context of the platform, maybe they expect something like:

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

But since I can't be sure, and given that the original problem is likely from a competition where multiple answers are boxed separately, I think that's acceptable. However, to comply with the instruction "put your final answer within \boxed{}", maybe they want all answers in one box. Given that, I will go with:

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

Yes, that seems to fit. 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. A key insight is to recognize that each numerator is of the form $ (x + a)(x - b) $, and the denominator is a multiple of $ (x + a)(x - b) $, allowing us to rewrite each term in a more manageable form.

---

### Step 1: Simplify Each Term

Let’s define:

- $ D_1 = 5(x + 2)(x - 4) $
- $ D_2 = 9(x + 4)(x - 6) $
- $ D_3 = 13(x + 6)(x - 8) $

Then, the original equation becomes:

$$
\frac{(x+1)(x-3)}{D_1} + \frac{(x+3)(x-5)}{D_2} - \frac{2(x+5)(x-7)}{D_3} = \frac{92}{585}
$$

We observe that each numerator is of the form $ (x + a)(x - b) $, and the denominator is a multiple of the same expression. This allows us to rewrite each term as:

- $ \frac{(x+1)(x-3)}{D_1} = \frac{1}{5} + \frac{5}{D_1} $
- $ \frac{(x+3)(x-5)}{D_2} = \frac{1}{9} + \frac{9}{D_2} $
- $ \frac{2(x+5)(x-7)}{D_3} = \frac{2}{13} + \frac{26}{D_3} $

So the equation becomes:

$$
\left( \frac{1}{5} + \frac{1}{9} - \frac{2}{13} \right) + \left( \frac{5}{D_1} + \frac{9}{D_2} - \frac{26}{D_3} \right) = \frac{92}{585}
$$

---

### Step 2: Compute the Constant Term

We compute:

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

Find a common denominator, which is 585:

$$
\frac{1}{5} = \frac{117}{585}, \quad \frac{1}{9} = \frac{65}{585}, \quad \frac{2}{13} = \frac{90}{585}
$$

So:

$$
\frac{117 + 65 - 90}{585} = \frac{92}{585}
$$

Thus, the equation simplifies to:

$$
\frac{5}{D_1} + \frac{9}{D_2} - \frac{26}{D_3} = 0
$$

---

### Step 3: Substitute and Simplify

Recall:

- $ D_1 = 5(x + 2)(x - 4) \Rightarrow \frac{5}{D_1} = \frac{1}{(x + 2)(x - 4)} $
- $ D_2 = 9(x + 4)(x - 6) \Rightarrow \frac{9}{D_2} = \frac{1}{(x + 4)(x - 6)} $
- $ D_3 = 13(x + 6)(x - 8) \Rightarrow \frac{26}{D_3} = \frac{2}{(x + 6)(x - 8)} $

So the equation becomes:

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

---

### Step 4: Use Substitution

Let $ y = x^2 - 2x $. Then:

- $ (x + 2)(x - 4) = x^2 - 2x - 8 = y - 8 $
- $ (x + 4)(x - 6) = x^2 - 2x - 24 = y - 24 $
- $ (x + 6)(x - 8) = x^2 - 2x - 48 = y - 48 $

So the equation becomes:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} - \frac{2}{y - 48} = 0
$$

---

### Step 5: Solve the Simplified Equation

Combine the first two terms:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} = \frac{(y - 24) + (y - 8)}{(y - 8)(y - 24)} = \frac{2y - 32}{(y - 8)(y - 24)}
$$

So the equation becomes:

$$
\frac{2y - 32}{(y - 8)(y - 24)} - \frac{2}{y - 48} = 0
$$

Factor out 2:

$$
2\left( \frac{y - 16}{(y - 8)(y - 24)} - \frac{1}{y - 48} \right) = 0
$$

So:

$$
\frac{y - 16}{(y - 8)(y - 24)} = \frac{1}{y - 48}
$$

Cross-multiplying:

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

Expand both sides:

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

Set equal:

$$
y^2 - 64y + 768 = y^2 - 32y + 192
$$

Subtract $ y^2 $ from both sides:

$$
-64y + 768 = -32y + 192
$$

Solve:

$$
-32y = -576 \Rightarrow y = 18
$$

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

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

Solve using the quadratic formula:

$$
x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-18)}}{2(1)} = \frac{2 \pm \sqrt{4 + 72}}{2} = \frac{2 \pm \sqrt{76}}{2} = 1 \pm \sqrt{19}
$$

---

### Step 6: Final Answer

Both solutions are valid and do not make any denominator zero in the original equation. Therefore, the real roots are:

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

[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 expanding didn't help. Maybe I can write each term as something over (x + something)(x - something). Let me think.

Alternatively, maybe notice that each term has a similar structure. Let me check:

First term numerator: (x + 1)(x - 3) = x² - 2x - 3. Denominator: 5(x + 2)(x - 4) = 5(x² - 2x - 8). So if I write numerator as (x² - 2x - 8) + 5. Because x² - 2x - 8 + 5 = x² - 2x - 3. So numerator is denominator/5 + 5? Wait:

Let me check:

Denominator is 5(x² - 2x - 8). So if I take denominator divided by 5, that's x² - 2x - 8. Then adding 5 gives x² - 2x - 8 + 5 = x² - 2x - 3, which is the numerator. So numerator = (denominator / 5) + 5. Therefore, the first term can be written as:

[ (denominator / 5) + 5 ] / denominator = (denominator / 5)/denominator + 5 / denominator = 1/5 + 5 / denominator.

Wait, that seems useful! Let me verify:

Let me denote D1 = 5(x + 2)(x - 4). Then numerator is (x + 1)(x - 3) = D1 / 5 + 5? Let me check:

D1 / 5 = (x + 2)(x - 4) = x² - 2x - 8. Then adding 5 gives x² - 2x - 8 + 5 = x² - 2x - 3, which is the numerator. Yes! So the first term is [D1 / 5 + 5]/D1 = 1/5 + 5/D1.

Similarly, maybe the same approach can be applied to the other terms?

Let me check the second term: numerator is (x + 3)(x - 5) = x² - 2x -15. Denominator is 9(x + 4)(x - 6) = 9(x² - 2x - 24). Let me check if numerator can be expressed as denominator / 9 + something.

Denominator / 9 = (x + 4)(x - 6) = x² - 2x - 24. Then adding something to get numerator x² - 2x -15. So x² - 2x -24 + something = x² - 2x -15. Therefore, something is 9. So numerator = denominator / 9 + 9. Therefore, the second term is [denominator / 9 + 9]/denominator = 1/9 + 9 / denominator.

Similarly for the third term: numerator is 2(x + 5)(x - 7) = 2(x² - 2x - 35). Denominator is 13(x + 6)(x - 8) = 13(x² - 2x - 48). Let me check if numerator can be expressed as denominator / 13 + something.

Denominator / 13 = (x + 6)(x - 8) = x² - 2x - 48. Then adding something to get numerator 2(x² - 2x - 35) = 2x² - 4x -70. Wait, but denominator / 13 is x² - 2x - 48. So if I multiply denominator / 13 by 2, I get 2x² - 4x - 96. Then adding something to get 2x² -4x -70. So 2x² -4x -96 + something = 2x² -4x -70. Therefore, something is 26. Therefore, numerator = 2*(denominator / 13) + 26. Therefore, the third term is [2*(denominator / 13) + 26]/denominator = 2/13 + 26 / denominator.

Wait, but the third term is subtracted. Let me check:

Original third term is - [2(x + 5)(x - 7)] / [13(x + 6)(x - 8)] = - [2*(denominator / 13) + 26]/denominator = - [2/13 + 26 / denominator]. So that would be -2/13 - 26 / denominator.

So putting all together, the original equation becomes:

First term: 1/5 + 5 / D1

Second term: 1/9 + 9 / D2

Third term: -2/13 - 26 / D3

So adding them all up:

(1/5 + 1/9 - 2/13) + (5/D1 + 9/D2 - 26/D3) = 92/585.

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

To compute that, find common denominators. Let me compute:

First, 1/5 + 1/9 = (9 + 5)/45 = 14/45.

Then subtract 2/13: 14/45 - 2/13. Let me compute that:

Find common denominator of 45 and 13, which is 585.

Convert 14/45 to (14 * 13)/585 = 182/585.

Convert 2/13 to (2 * 45)/585 = 90/585.

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

Wait! So the constants add up to 92/585. Therefore, the equation becomes:

92/585 + (5/D1 + 9/D2 - 26/D3) = 92/585.

Subtracting 92/585 from both sides:

5/D1 + 9/D2 - 26/D3 = 0.

So now, the equation simplifies to:

5/D1 + 9/D2 - 26/D3 = 0.

Where D1 = 5(x + 2)(x - 4), D2 = 9(x + 4)(x - 6), D3 = 13(x + 6)(x - 8).

Wait, but let me check:

Earlier, we had:

First term: 1/5 + 5/D1

Second term: 1/9 + 9/D2

Third term: -2/13 - 26/D3

So adding them up:

(1/5 + 1/9 - 2/13) + (5/D1 + 9/D2 - 26/D3) = 92/585.

But we found that (1/5 + 1/9 - 2/13) = 92/585, so the equation reduces to:

92/585 + (5/D1 + 9/D2 - 26/D3) = 92/585 => 5/D1 + 9/D2 - 26/D3 = 0.

So now, the problem reduces to solving:

5/D1 + 9/D2 - 26/D3 = 0.

But D1, D2, D3 are denominators from original terms. Let me write them again:

D1 = 5(x + 2)(x - 4)

D2 = 9(x + 4)(x - 6)

D3 = 13(x + 6)(x - 8)

So substituting:

5 / [5(x + 2)(x - 4)] + 9 / [9(x + 4)(x - 6)] - 26 / [13(x + 6)(x - 8)] = 0.

Simplify each term:

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

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

Third term: -26 / [13(x + 6)(x - 8)] = -2 / [(x + 6)(x - 8)]

So the equation simplifies to:

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

That's a big simplification! So now, the equation is:

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

This seems more manageable. Let me write this equation again:

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

Now, perhaps I can combine these fractions. To do that, I need a common denominator. However, the denominators are all quadratic terms. Let me note that each denominator is of the form (x + a)(x - b), where a and b are constants. Let me check if there's a pattern or substitution.

Alternatively, maybe I can use partial fractions or some substitution. Let me think. Alternatively, notice that each denominator is similar to the previous one with shifted terms. Let me check:

First denominator: (x + 2)(x - 4) = x² - 2x - 8

Second denominator: (x + 4)(x - 6) = x² - 2x - 24

Third denominator: (x + 6)(x - 8) = x² - 2x - 48

Wait, interesting! Each denominator is x² - 2x - something. Let me denote y = x² - 2x. Then denominators are y - 8, y - 24, y - 48. So maybe substitution?

Let me try that. Let me set y = x² - 2x. Then denominators become:

First: y - 8

Second: y - 24

Third: y - 48

So the equation becomes:

1/(y - 8) + 1/(y - 24) - 2/(y - 48) = 0.

That seems promising! Let me write that:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} - \frac{2}{y - 48} = 0
$$

Now, solving for y. Let me combine these terms. Let me first combine the first two terms:

1/(y - 8) + 1/(y - 24) = [ (y - 24) + (y - 8) ] / [(y - 8)(y - 24)] = [2y - 32]/[(y - 8)(y - 24)].

So the equation becomes:

[2y - 32]/[(y - 8)(y - 24)] - 2/(y - 48) = 0.

Let me write that as:

[2(y - 16)] / [(y - 8)(y - 24)] - 2/(y - 48) = 0.

Factor out the 2:

2[ (y - 16)/[(y - 8)(y - 24)] - 1/(y - 48) ] = 0.

Since 2 ≠ 0, we can divide both sides by 2:

(y - 16)/[(y - 8)(y - 24)] - 1/(y - 48) = 0.

So:

(y - 16)/[(y - 8)(y - 24)] = 1/(y - 48).

Cross-multiplying:

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

Let me expand both sides.

Left side: (y - 16)(y - 48) = y² - 48y -16y + 16*48 = y² - 64y + 768.

Right side: (y - 8)(y - 24) = y² -24y -8y + 192 = y² -32y + 192.

Set left side equal to right side:

y² - 64y + 768 = y² -32y + 192.

Subtract y² from both sides:

-64y + 768 = -32y + 192.

Bring all terms to left side:

-64y + 768 + 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, we need to check if these solutions are valid, i.e., they don't make any denominators zero in the original equation.

Original denominators in the simplified equation were (x + 2)(x - 4), (x + 4)(x - 6), (x + 6)(x - 8). Let me check for x = 1 ± sqrt(19).

First, compute approximate values to check if they are in the domain.

sqrt(19) is approx 4.3589.

So x = 1 + 4.3589 ≈ 5.3589

x = 1 - 4.3589 ≈ -3.3589

Check denominators for x ≈ 5.3589:

First denominator: (x + 2)(x - 4) ≈ (5.3589 + 2)(5.3589 - 4) ≈ (7.3589)(1.3589) ≈ not zero.

Second denominator: (x + 4)(x - 6) ≈ (5.3589 + 4)(5.3589 - 6) ≈ (9.3589)(-0.6411) ≈ not zero.

Third denominator: (x + 6)(x - 8) ≈ (5.3589 + 6)(5.3589 - 8) ≈ (11.3589)(-2.6411) ≈ not zero.

Similarly for x ≈ -3.3589:

First denominator: (x + 2)(x - 4) ≈ (-3.3589 + 2)(-3.3589 - 4) ≈ (-1.3589)(-7.3589) ≈ not zero.

Second denominator: (x + 4)(x - 6) ≈ (-3.3589 + 4)(-3.3589 - 6) ≈ (0.6411)(-9.3589) ≈ not zero.

Third denominator: (x + 6)(x - 8) ≈ (-3.3589 + 6)(-3.3589 - 8) ≈ (2.6411)(-11.3589) ≈ not zero.

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

But let me check if there are any other solutions. Wait, when we did the substitution, we had y = x² - 2x, and solved for y = 18, leading to quadratic equation. So only two solutions. Therefore, these are the only real roots.

But let me check if there was any step where we might have introduced extraneous solutions or lost solutions. For example, when we cross-multiplied, we assumed that denominators are not zero. But we already checked that our solutions don't make denominators zero, so they are valid.

Alternatively, let me check by plugging back into the original equation. Let me take x = 1 + sqrt(19). Let me compute the left-hand side (LHS) and see if it equals 92/585.

But that might be tedious. Alternatively, since we followed algebraic steps correctly, and checked that the solutions don't make denominators zero, they should be valid. However, to be thorough, let me check with approximate values.

First, compute approximate value of sqrt(19) ≈ 4.3589, so x ≈ 5.3589.

Compute each term:

First term: [(x+1)(x-3)] / [5(x+2)(x-4)]

Compute numerator: (5.3589 + 1)(5.3589 - 3) ≈ (6.3589)(2.3589) ≈ approx 6.3589 * 2.3589 ≈ let me compute 6 * 2.3589 = 14.1534, 0.3589 * 2.3589 ≈ approx 0.846, total ≈ 14.1534 + 0.846 ≈ 14.999 ≈ 15.

Denominator: 5*(5.3589 + 2)*(5.3589 - 4) = 5*(7.3589)*(1.3589) ≈ 5*(7.3589*1.3589). Let me compute 7 * 1.3589 ≈ 9.5123, 0.3589*1.3589 ≈ approx 0.487, total ≈ 9.5123 + 0.487 ≈ 10. So denominator ≈ 5*10 = 50. Therefore, first term ≈ 15 / 50 = 0.3.

Second term: [(x+3)(x-5)] / [9(x+4)(x-6)]

Numerator: (5.3589 + 3)(5.3589 - 5) ≈ (8.3589)(0.3589) ≈ approx 8 * 0.3589 = 2.8712 + 0.3589*0.3589 ≈ approx 2.8712 + 0.128 ≈ 3.0.

Denominator: 9*(5.3589 + 4)*(5.3589 - 6) = 9*(9.3589)*(-0.6411) ≈ 9*( -5.999 ) ≈ -53.991. Therefore, second term ≈ 3 / (-53.991) ≈ -0.0556.

Third term: - [2(x+5)(x-7)] / [13(x+6)(x-8)]

Numerator: 2*(5.3589 + 5)*(5.3589 - 7) = 2*(10.3589)*(-1.6411) ≈ 2*(-16.98) ≈ -33.96.

Denominator: 13*(5.3589 + 6)*(5.3589 - 8) = 13*(11.3589)*(-2.6411) ≈ 13*(-29.96) ≈ -389.48.

Third term ≈ - [ -33.96 / -389.48 ] = - [ 0.0872 ] ≈ -0.0872.

So total LHS ≈ 0.3 - 0.0556 - 0.0872 ≈ 0.3 - 0.1428 ≈ 0.1572.

Now, 92/585 ≈ let me compute that. 585 divided by 92: 585 ÷ 92 ≈ 6.3587. So 1/6.3587 ≈ 0.1572. Therefore, 92/585 ≈ 0.1572. Which matches the approximate value of LHS. Therefore, x ≈ 5.3589 is a valid solution.

Similarly, check x ≈ -3.3589.

First term: [(x+1)(x-3)] / [5(x+2)(x-4)]

Numerator: (-3.3589 + 1)(-3.3589 - 3) = (-2.3589)(-6.3589) ≈ 15.

Denominator: 5*(-3.3589 + 2)*(-3.3589 - 4) = 5*(-1.3589)*(-7.3589) ≈ 5*(10) = 50. So first term ≈ 15 / 50 = 0.3.

Second term: [(x+3)(x-5)] / [9(x+4)(x-6)]

Numerator: (-3.3589 + 3)(-3.3589 - 5) = (-0.3589)(-8.3589) ≈ 2.999 ≈ 3.

Denominator: 9*(-3.3589 + 4)*(-3.3589 - 6) = 9*(0.6411)*(-9.3589) ≈ 9*(-5.999) ≈ -53.991. So second term ≈ 3 / (-53.991) ≈ -0.0556.

Third term: - [2(x+5)(x-7)] / [13(x+6)(x-8)]

Numerator: 2*(-3.3589 + 5)*(-3.3589 - 7) = 2*(1.6411)*(-10.3589) ≈ 2*(-16.98) ≈ -33.96.

Denominator: 13*(-3.3589 + 6)*(-3.3589 - 8) = 13*(2.6411)*(-11.3589) ≈ 13*(-29.96) ≈ -389.48.

Third term ≈ - [ -33.96 / -389.48 ] = - [ 0.0872 ] ≈ -0.0872.

So total LHS ≈ 0.3 - 0.0556 - 0.0872 ≈ same as before, 0.1572 ≈ 92/585. Therefore, x ≈ -3.3589 is also a valid solution.

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

But let me check if there are any other solutions. For example, when we did substitution y = x² - 2x, and solved for y = 18, leading to quadratic equation. So only two solutions. Therefore, these are the only real roots.

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

But the problem says "Enter the real roots, separated by commas." So maybe as \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}? Or maybe in a single box? Wait, the original instruction says "Enter the real roots, separated by commas." So likely, they want both roots in one box, separated by commas. Let me check the problem statement again:

"Enter the real roots, separated by commas."

So probably, the answer should be written as \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}}? Or maybe inside a single box? But standard practice for multiple answers is to put them in separate boxes or in one box separated by commas. Since the user instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the original problem says "Enter the real roots, separated by commas." So maybe inside a single box, separated by commas. Let me check previous problems. For example, if there are multiple answers, they are often put in one box separated by commas. For example, \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} might be two boxes, but maybe the user wants them in one box. However, since the user says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the instruction is ambiguous. However, given that the original equation is likely to have two real roots, and the standard practice here is to put multiple answers in separate boxes. However, since the user might have intended a single box, but I need to check.

Alternatively, maybe I can check if there are more solutions. Wait, we had y = x² - 2x = 18, leading to x² - 2x - 18 = 0, which has two real roots. So only two real roots. Therefore, the answer is those two. So I think the correct way is to write both in boxed form separated by commas. However, since the user says "put your final answer within \boxed{}", maybe they want both answers in one box. For example:

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

But maybe the system expects multiple boxed answers. However, since the original problem is from an Olympiad or similar, and the standard is to write all real roots separated by commas inside a single box. For example, like \boxed{1 + \sqrt{19}}, \boxed{1 - \sqrt{19}} or \boxed{1 + \sqrt{19}, 1 - \sqrt{19}}.

But I need to check the exact instruction. The user says: "Enter the real roots, separated by commas." and "put your final answer within \boxed{}". So maybe they want both roots inside a single box, separated by commas. For example:

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

But I need to check if that's acceptable. In some contexts, multiple answers are put in separate boxes, but since the user says "within \boxed{}", maybe they want one box. However, the exact instruction is ambiguous. However, given that the original equation is likely to have two real roots, and the standard practice in many competitions is to list all solutions in a single box separated by commas, I will go with that.

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

Wait, but the user might expect both in one box. Let me check again. The original instruction says: "Enter the real roots, separated by commas." So maybe:

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

But I think the standard on platforms like AoPS or others is to put multiple answers in separate boxes. However, since the user hasn't specified, but given the initial problem statement, I think safest is to put both in one box separated by commas. However, in many cases, multiple boxed answers are acceptable. But since the user says "put your final answer within \boxed{}", maybe they want a single box. Let me check the problem again. The original equation is a single equation, so the answer is two real roots. In many cases, for such problems, the answer is written as two boxed expressions. However, since the user might be using an automated system that expects a single box, but I can't be sure. Given that, I think safest is to follow the instruction: "Enter the real roots, separated by commas." So if I have to write them in one box, separated by commas. So:

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

But I need to check if that's acceptable. In some contexts, commas inside the box are okay. For example, on AoPS, they sometimes do that. Alternatively, if the system is expecting multiple boxed answers, but since the user says "put your final answer within \boxed{}", maybe they want one box. Given that, I will go with:

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

But since the user might have intended a single box, but I can't be sure. However, given that the original instruction says "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will write both roots inside a single box separated by commas.

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

Wait, but that's two boxes. Maybe the user wants them in one box. Let me check the exact instruction again: "Enter the real roots, separated by commas." So if I have to enter them separated by commas, then in the context of the platform, maybe they expect something like:

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

But since I can't be sure, and given that the original problem is likely from a competition where multiple answers are boxed separately, I think that's acceptable. However, to comply with the instruction "put your final answer within \boxed{}", maybe they want all answers in one box. Given that, I will go with:

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

Yes, that seems to fit. 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. A key insight is to recognize that each numerator is of the form $ (x + a)(x - b) $, and the denominator is a multiple of $ (x + a)(x - b) $, allowing us to rewrite each term in a more manageable form.

---

### Step 1: Simplify Each Term

Let’s define:

- $ D_1 = 5(x + 2)(x - 4) $
- $ D_2 = 9(x + 4)(x - 6) $
- $ D_3 = 13(x + 6)(x - 8) $

Then, the original equation becomes:

$$
\frac{(x+1)(x-3)}{D_1} + \frac{(x+3)(x-5)}{D_2} - \frac{2(x+5)(x-7)}{D_3} = \frac{92}{585}
$$

We observe that each numerator is of the form $ (x + a)(x - b) $, and the denominator is a multiple of the same expression. This allows us to rewrite each term as:

- $ \frac{(x+1)(x-3)}{D_1} = \frac{1}{5} + \frac{5}{D_1} $
- $ \frac{(x+3)(x-5)}{D_2} = \frac{1}{9} + \frac{9}{D_2} $
- $ \frac{2(x+5)(x-7)}{D_3} = \frac{2}{13} + \frac{26}{D_3} $

So the equation becomes:

$$
\left( \frac{1}{5} + \frac{1}{9} - \frac{2}{13} \right) + \left( \frac{5}{D_1} + \frac{9}{D_2} - \frac{26}{D_3} \right) = \frac{92}{585}
$$

---

### Step 2: Compute the Constant Term

We compute:

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

Find a common denominator, which is 585:

$$
\frac{1}{5} = \frac{117}{585}, \quad \frac{1}{9} = \frac{65}{585}, \quad \frac{2}{13} = \frac{90}{585}
$$

So:

$$
\frac{117 + 65 - 90}{585} = \frac{92}{585}
$$

Thus, the equation simplifies to:

$$
\frac{5}{D_1} + \frac{9}{D_2} - \frac{26}{D_3} = 0
$$

---

### Step 3: Substitute and Simplify

Recall:

- $ D_1 = 5(x + 2)(x - 4) \Rightarrow \frac{5}{D_1} = \frac{1}{(x + 2)(x - 4)} $
- $ D_2 = 9(x + 4)(x - 6) \Rightarrow \frac{9}{D_2} = \frac{1}{(x + 4)(x - 6)} $
- $ D_3 = 13(x + 6)(x - 8) \Rightarrow \frac{26}{D_3} = \frac{2}{(x + 6)(x - 8)} $

So the equation becomes:

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

---

### Step 4: Use Substitution

Let $ y = x^2 - 2x $. Then:

- $ (x + 2)(x - 4) = x^2 - 2x - 8 = y - 8 $
- $ (x + 4)(x - 6) = x^2 - 2x - 24 = y - 24 $
- $ (x + 6)(x - 8) = x^2 - 2x - 48 = y - 48 $

So the equation becomes:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} - \frac{2}{y - 48} = 0
$$

---

### Step 5: Solve the Simplified Equation

Combine the first two terms:

$$
\frac{1}{y - 8} + \frac{1}{y - 24} = \frac{(y - 24) + (y - 8)}{(y - 8)(y - 24)} = \frac{2y - 32}{(y - 8)(y - 24)}
$$

So the equation becomes:

$$
\frac{2y - 32}{(y - 8)(y - 24)} - \frac{2}{y - 48} = 0
$$

Factor out 2:

$$
2\left( \frac{y - 16}{(y - 8)(y - 24)} - \frac{1}{y - 48} \right) = 0
$$

So:

$$
\frac{y - 16}{(y - 8)(y - 24)} = \frac{1}{y - 48}
$$

Cross-multiplying:

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

Expand both sides:

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

Set equal:

$$
y^2 - 64y + 768 = y^2 - 32y + 192
$$

Subtract $ y^2 $ from both sides:

$$
-64y + 768 = -32y + 192
$$

Solve:

$$
-32y = -576 \Rightarrow y = 18
$$

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

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

Solve using the quadratic formula:

$$
x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-18)}}{2(1)} = \frac{2 \pm \sqrt{4 + 72}}{2} = \frac{2 \pm \sqrt{76}}{2} = 1 \pm \sqrt{19}
$$

---

### Step 6: Final Answer

Both solutions are valid and do not make any denominator zero in the original equation. Therefore, the real roots are:

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

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.)