Đáp án:
Cách 1:
$\frac{y}{2x²-xy}$ +$\frac{4x}{y²-2xy}$
= $\frac{y}{x(2x-y)}$ +$\frac{4x}{y(y-2x)}$
= $\frac{y}{x(2x-y)}$ + $\frac{-4x}{y(2x-y)}$
= $\frac{y.y}{xy(2x-y)}$+$\frac{-4x.x}{xy(2x-y)}$
= $\frac{y²-4x²}{xy(2x-y)}$
=$\frac{(y-2x)(y+2x)}{xy(2x-y)}$
=$\frac{-1.(y+2x)}{xy}$
=$\frac{-2x-y}{xy}$
Cách 2:
$\frac{y}{2x²-xy}$ +$\frac{4x}{y²-2xy}$
=$\frac{y}{2x²-xy}$-$\frac{4x}{-(y²-2xy)}$
=$\frac{y}{2x²-xy}$-$\frac{4x}{2xy-y²}$
=$\frac{y}{x(2x-y)}$-$\frac{4x}{y(2x-y)}$
=$\frac{y.y}{xy(2x-y)}$-$\frac{4x.x}{xy(2x-y}$
=$\frac{y²-4x²}{xy(2x-y)}$
=$\frac{(y-2x)(y+2x)}{xy(2x-y)}$
=$\frac{-1.(y+2x)}{xy}$
=$\frac{-2x-y}{xy}$
