Đáp án:
$\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}-\dfrac{1}{x+2}$
$=\dfrac{x^2}{x(x^2-4)}+\dfrac{6}{3(2-x)}-\dfrac{1}{x+2}$
$=\dfrac{x^2}{x(x-2)(x+2)}-\dfrac{6}{3(x-2)}-\dfrac{1}{x+2}$
$=\dfrac{3x^2-6x^2-12x-3x^2+6x}{3x(x-2)(x+2)}$
$=\dfrac{-6x^2-6x}{3x(x-2)(x+2)}$
$=\dfrac{-6x(x+1)}{3x(x-2)(x+2)}$
$=\dfrac{-2(x+1)}{(x-2)(x+2)}$
