Question

Reduce the fraction: x-y/x^2-1 times x-1/x^2-y^2

Answer 1

Answer:

l

Step-by-step explanation:

Answer 2

Answer: $\frac{1}{(x+1)(x+y)}$

Step-by-step explanation:

Given the expression:

$(\frac{x-y}{x^2-1})(\frac{x-1}{x^2-y^2})$

The first step is to multiply the numerator of the first fraction by the numerator of the second fraction and the denominator of the first fraction by the denominator of the second fraction. Then:

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

Since $(x^2-1)$ and $(x^2-y^2)$ are perfect squares, you can factorize them in the form:

$a^2-b^2=(a+b)(a-b)$

Then:

$=\frac{(x-y)(x-1)}{(x+1)(x-1)(x+y)(x-y)}$

Simplifying, you get:

$=\frac{1}{(x+1)(x+y)}$