Giải thích các bước giải:

a.Ta có:

$\dfrac{2x^2y^5}{3x^4y^2}=\dfrac{2y^3}{3x^2}$

b.Ta có:

$\dfrac{3x(x-y)^3}{2x^2(x-y)^2}=\dfrac{3(x-y)}{2x}$

c.Ta có:

$\dfrac{3x^2y+4xy^2}{6x+8y}=\dfrac{xy(3x+4y)}{2(3x+4y)}=\dfrac{xy}2$

d.Ta có:

$\dfrac{-3x^2-6x}{4-x^2}=\dfrac{-3x(x+2)}{(2-x)(2+x)}=\dfrac{-3x}{2-x}=\dfrac{3x}{x-2}$