Respuesta :

kudzordzifrancis

Answer:

[tex](\frac{1}{(x+1)(x+y)})[/tex]

Step-by-step explanation:

The given algebraic fraction is;

[tex](\frac{x-y}{x^2-1})(\frac{x-1}{x^2-y^2} )[/tex]

We factor using difference of two squares.

[tex](\frac{x-y}{x^2-1^2})(\frac{x-1}{x^2-y^2} )[/tex]

[tex](\frac{x-y}{(x+1)(x-1)})(\frac{x-1}{(x-y)(x+y)} )[/tex]

Cancel out the common factors;

[tex](\frac{1}{(x+1)(1)})(\frac{1}{(1)(x+y)} )[/tex]

Simplify;

[tex](\frac{1}{(x+1)(x+y)})[/tex]

zainebamir540

Answer:

1/(x+1)(x+y) is the simplification of (x-y/x²-1)(x-1/x²-y²).

Step-by-step explanation:

We have given the expression:

(x-y/x²-1)(x-1/x²-y²)

We know that:

a²-b² = (a-b)(a+b) we get,

(x-y/(x+1)(x-1)) × ((x-1)/(x-y)(x+y))

We cancel out the like terms we get,

(1/x+1)×(1/x+y)

Simplification is :

1/(x+1)(x+y) is the simplification of (x-y/x²-1)(x-1/x²-y²).

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