Question

решите пж по действиям
$( \frac{x + 1}{x - 1} - \frac{x - 1}{x +1 }) - \frac{4x}{ {x}^{2} - 1 } =$
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Answer 1

Ответ:

$=0$

Объяснение:

$\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)-\frac{4x}{x^2-1}$

$=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2-1}$

$=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{\left(x+1\right)\left(x-1\right)}$

$=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{4x}{\left(x-1\right)\left(x+1\right)}$

$=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x}{\left(x-1\right)\left(x+1\right)}$

$=\frac{0}{\left(x-1\right)\left(x+1\right)}$

$=0$