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IuliusCaesar

[tex]\frac{x-y}{x^{2} -1}[/tex]*[tex]\frac{x-1}{x^{2}-y^{2}}[/tex]=

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

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

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

abidemiokin

Answer:

The answer is 1/(x+1)(x+y)

Step-by-step explanation:

Given the equation

x-y/x²-1 × x-1/x²-y²

We will apply the difference of two square to solve the equation. According to the rule, if a and b are real numbers;

a²-b² = (a+b)(a-b)

Therefore x²-1 = x²-1² = (x+1)(x-1)

Similarly, x²-y² = (x+y)(x-y)

Substituting this functions in the question we have;

x-y/x²-1 × x-1/x²-y²

= x-y/(x+1)(x-1) × x-1/(x+y)(x-y)

x-y will cancel x-y and x-1 will also cancel out x-1 to reduce the equation to;

x-y/x²-1 × x-1/x²-y² = 1/x+1 × 1/x+y

= 1/(x+1)(x+y) which is the simplified form of the equation

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