To solve the equation

\[ \frac{x+5}{x-5} - \frac{x-5}{x+5} = \frac{20}{x^2-25}, \]

we first recognize that \( x^2 - 25 \) can be factored as \( (x-5)(x+5) \). This allows us to rewrite the equation with a common denominator.

First, let's rewrite the left-hand side with a common denominator:

\[ \frac{x+5}{x-5} - \frac{x-5}{x+5} = \frac{(x+5)(x+5) - (x-5)(x-5)}{(x-5)(x+5)}. \]

Simplify the numerators:

\[ (x+5)(x+5) = x^2 + 10x + 25, \]
\[ (x-5)(x-5) = x^2 - 10x + 25. \]

So the equation becomes:

\[ \frac{x^2 + 10x + 25 - (x^2 - 10x + 25)}{(x-5)(x+5)} = \frac{20}{(x-5)(x+5)}. \]

Simplify the numerator:

\[ x^2 + 10x + 25 - x^2 + 10x - 25 = 20x. \]

Thus, the equation is:

\[ \frac{20x}{(x-5)(x+5)} = \frac{20}{(x-5)(x+5)}. \]

Since the denominators are the same, we can equate the numerators:

\[ 20x = 20. \]

Solving for \( x \):

\[ x = 1. \]

Finally, we should check if \( x = 1 \) is a valid solution by substituting it back into the original equation to ensure it does not make any denominator zero:

\[ \frac{1+5}{1-5} - \frac{1-5}{1+5} = \frac{20}{1^2-25}. \]

Simplify:

\[ \frac{6}{-4} - \frac{-4}{6} = \frac{20}{-24}, \]
\[ -\frac{3}{2} + \frac{2}{3} = -\frac{5}{6}. \]

Both sides simplify to \(-\frac{5}{6}\), confirming that \( x = 1 \) is indeed a solution.