Question

$[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}]÷\frac{4}{x^4-1}$, when simplified is equal to

• A. 1
• B. 0
• C. $x^2-1$
• D. 2x

Answer 1

The correct option is D 2x

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