Đáp án:
$(\dfrac{x}{x^2-4}+\dfrac{1}{x-2}):(\dfrac{1}{x-2}-1)$$=\dfrac{2x+2}{(x+2)(-x+3)}$
Giải thích các bước giải:
$(\dfrac{x}{x^2-4}+\dfrac{1}{x-2}):(\dfrac{1}{x-2}-1)$
$=(\dfrac{x}{(x-2)(x+2)}+\dfrac{x+2}{(x-2)(x+2)}):\dfrac{1-(x-2)}{x-2}$
$=\dfrac{x+x+2}{(x-2)(x+2)}:\dfrac{-x+3}{x-2}$
$=\dfrac{2x+2}{(x-2)(x+2)}.\dfrac{x-2}{-x+3}$
$=\dfrac{2x+2}{(x+2)(-x+3)}$