Đáp án:
C= ($\dfrac{x+2}{2-x}$ + $\dfrac{4x^2}{4-x^2}$ + $\dfrac{2-x}{x+2}$ ) : $\dfrac{x^2-x}{2x-x^2}$
ĐKXĐ : x $\neq$ ± 2 ; x $\neq$ 0
= ( $\dfrac{(x+2)² + 4x^2 + (2-x)²}{(2-x)(2+x)}$) : $\dfrac{x(x-1)}{x(2-x)}$
= $\dfrac{x²+4x+4 +4x² + 4-4x+x²}{(2-x)(2+x)}$ . $\dfrac{x(2-x)}{x(x-1)}$
= $\dfrac{6x²+8}{(2-x)(2+x)}$ . $\dfrac{(2-x)}{(x-1)}$
= $\dfrac{6x²+8}{(2+x)(x-1)}$