Đáp án:
$1$
Giải thích các bước giải:
$ĐK : x \neq ± 2$
$(\frac{x}{x - 2}+\frac{2}{x + 2}-\frac{8- 8x}{4-x²}):\frac{x-2}{xx+2}$
$= (\frac{x}{x - 2} + \frac{2}{x + 2}+ \frac{8-8x}{(x+2)(x-2)}):\frac{x- 2}{x + 2}$
$= \frac{x(x + 2)+ 2(x - 2)+8-8x}{(x+2)(x-2)}.\frac{x+2}{x - 2}$
$=\frac{x² + 2x + 2x - 4 + 8- 8x}{(x +2)(x-2)}.\frac{x+2}{x-2}$
$=\frac{x² -4x + 4}{(x+2)(x-2)}.\frac{x+2}{x-2}$
$= \frac{(x-2)²}{(x+2)(x-2)}.\frac{x+2}{x - 2}$
$= \frac{(x-2)²(x+2)}{(x+2)(x-2)²} = 1$
