Đáp án:
Giải thích các bước giải:
$MTC:y.(x-y)^3$
$f)\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}$
$=\dfrac{x^3}{(x-y)^3}$
$=\dfrac{x^3y}{(x-y)^3y}$
$+)\dfrac{x}{y^2-xy}$
$=\dfrac{x}{-y.(x-y)}$
$=\dfrac{x.(x-y)^2}{-y.(x-y).(x-y)^2}$
$=\dfrac{-x(x-y)^2}{y.(x-y)^3}$