We are asked to simplify two expressions. 1. $\frac{x^2-10x+25}{x^2-25}$

1. Problem Description

We are asked to simplify two expressions.

1. $\frac{x^2-10x+25}{x^2-25}$

2. $\frac{x^2-3x}{x^2-x-2} \times \frac{x-2}{x^2-6x+9}$

2. Solution Steps

Problem 1:

We can factor the numerator and the denominator.

The numerator is a perfect square trinomial.

$x^2 - 10x + 25 = (x-5)(x-5) = (x-5)^2$

The denominator is a difference of squares.

$x^2 - 25 = (x-5)(x+5)$

So the expression becomes

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

Problem 2:

We can factor each of the polynomials.

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

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

$x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2$

So the expression becomes

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

We can cancel the common factors of $(x-3)$ and $(x-2)$

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

3. Final Answer

1. $\frac{x-5}{x+5}$

2. $\frac{x}{(x+1)(x-3)}$ or $\frac{x}{x^2-2x-3}$