The problem is to simplify the expression: $\frac{2x-5}{x^2-1} + \frac{x^2-9}{x^2+3x-4} \times \frac{x^2-x-20}{x^2-2x-15}$.

1. Problem Description

The problem is to simplify the expression: $\frac{2x-5}{x^2-1} + \frac{x^2-9}{x^2+3x-4} \times \frac{x^2-x-20}{x^2-2x-15}$.

2. Solution Steps

First, we factorize the quadratic expressions:

$x^2 - 1 = (x-1)(x+1)$

$x^2 - 9 = (x-3)(x+3)$

$x^2 + 3x - 4 = (x+4)(x-1)$

$x^2 - x - 20 = (x-5)(x+4)$

$x^2 - 2x - 15 = (x-5)(x+3)$

Now, we substitute these factorizations into the expression:

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

We simplify the multiplication part:

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

We cancel out the common terms $(x+3)$, $(x+4)$, and $(x-5)$:

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

Now the expression becomes:

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

To add the two terms, we need a common denominator, which is $(x-1)(x+1)$:

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

Expand the term $(x-3)(x+1)$:

$(x-3)(x+1) = x^2 + x - 3x - 3 = x^2 - 2x - 3$

Substitute back into the expression:

$\frac{2x-5 + x^2 - 2x - 3}{(x-1)(x+1)} = \frac{x^2 - 8}{(x-1)(x+1)} = \frac{x^2 - 8}{x^2 - 1}$

3. Final Answer

$\frac{x^2-8}{x^2-1}$