Rút gọn biểu thức

$\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)$

$=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)$

$=\left(\frac{x}{x^2-4}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{6}{x+2}$

$=\left(\frac{x}{(x+2)(x-2)}-\frac{2(x+2)}{(x+2)(x-2)}+\frac{x-2}{(x-2)(x+2)}\right):\frac{6}{x+2}$

$=\frac{x-2x-4+x-2}{(x+2)(x-2)}.\frac{x+2}{6}$

$=\frac{-6}{(x+2)(x-2)}.\frac{x+2}{6}=\frac{1}{2-x}$

Đề như này mới ra đúng đáp án nhé