Question

Simplify: $[\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{8x}{x^2-16}]÷\frac{8x}{x^4-16}$

Answer 1

$[\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{8x}{x^2-16}]÷\frac{8x}{x^2-16}$

$=[\frac{x^2+4x+4-x^2+4x-4}{x^2-4}-\frac{8x}{x^2-16}]÷\frac{8x}{x^2-16}$

$=[\frac{8x}{x^2-4}-\frac{8x}{x^2-16}]÷\frac{8x}{x^2-16}$

$=[\frac{8x}{x^2-4}-\frac{8x}{(x^2-4)(x^2+4)}]÷\frac{8x}{x^2-16}$

$=\left[\frac{8x}{x^2-4}(1-\frac{1}{x^2+4})\right]÷\frac{8x}{x^2-16}$

$=[\frac{8x}{x^2-4}\times \frac{x^2+4-1}{x^2+4}]÷\frac{8x}{x^2-16}$

$=\left[\frac{8x(x^2+31)}{x^2-16}\right]÷\frac{8x}{x^2-16}$

$=\frac{8x(x^2+31)}{x^2-16}\times \frac{x^2-16}{8x}$

$=x^2+31$