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