Sure, let's simplify the expression step-by-step.
We are given the expression:
[tex]\[ \frac{x-y}{x^2-1} \cdot \frac{x-1}{x^2-y^2} \][/tex]
To simplify this expression, we will first factorize the denominators:
1. Factorize [tex]\(x^2 - 1\)[/tex] and [tex]\(x^2 - y^2\)[/tex]:
- [tex]\(x^2 - 1\)[/tex] is a difference of squares and can be factored as [tex]\((x-1)(x+1)\)[/tex].
- [tex]\(x^2 - y^2\)[/tex] is also a difference of squares and can be factored as [tex]\((x-y)(x+y)\)[/tex].
Given these factorizations, we can rewrite the expression as:
[tex]\[ \frac{x-y}{(x-1)(x+1)} \cdot \frac{x-1}{(x-y)(x+y)} \][/tex]
2. Cancel out common terms:
- We notice that [tex]\((x-y)\)[/tex] appears in the numerator of the first fraction and the denominator of the second fraction, so they can be cancelled out.
- Similarly, [tex]\((x-1)\)[/tex] appears in the numerator of the second fraction and the denominator of the first fraction, so they can be cancelled out as well.
Rewriting with the cancelled terms, we have:
[tex]\[ \frac{1}{(x+1)} \cdot \frac{1}{(x+y)} \][/tex]
3. Combine the fractions:
Since we are multiplying two fractions, we simply multiply the numerators together and the denominators together:
[tex]\[ \frac{1 \cdot 1}{(x+1)(x+y)} = \frac{1}{(x+1)(x+y)} \][/tex]
Finally, the simplified form of the given expression is:
[tex]\[ \frac{1}{x^2 + x y + x + y} \][/tex]
So, the answer is:
[tex]\[ \boxed{\frac{1}{x^2 + x y + x + y}} \][/tex]
