Giải thích các bước giải:
ĐKXĐ: $x\ge 0;x\ne 4;x\ne 9$
$C=\bigg{(}\dfrac{x+2\sqrt x}{x+4\sqrt x+4}+\dfrac{2x}{4-x}\bigg{)}:\bigg{(}\dfrac{\sqrt x-1}{x-2\sqrt x}-\dfrac{2\sqrt x+2}{x+\sqrt x}\bigg{)}$
$=\bigg{[}\dfrac{\sqrt x(\sqrt x+2)}{(\sqrt x+2)^2}-\dfrac{2x}{(\sqrt x-2)(\sqrt x+2)}\bigg{]}:\bigg{[}\dfrac{\sqrt x-1}{\sqrt x(\sqrt x-2)}-\dfrac{2(\sqrt x+1)}{\sqrt x(\sqrt x+1)}\bigg{]}$
$=\bigg{[}\dfrac{\sqrt x(\sqrt x-2)}{(\sqrt x+2)(\sqrt x-2)}-\dfrac{2x}{(\sqrt x-2)(\sqrt x+2)}\bigg{]}:\bigg{[}\dfrac{\sqrt x-1}{\sqrt x(\sqrt x-2)}-\dfrac{2(\sqrt x-2)}{\sqrt x(\sqrt x-2)}\bigg{]}$
$=\dfrac{x-2\sqrt x-2x}{(\sqrt x+2)(\sqrt x-2)}:\dfrac{\sqrt x-1-2\sqrt x+4}{\sqrt x(\sqrt x-2)}$
$=\dfrac{-\sqrt x(\sqrt x+2)}{(\sqrt x+2)(\sqrt x-2)}.\dfrac{\sqrt x(\sqrt x-2)}{3-\sqrt x}$
$=\dfrac{-x}{3-\sqrt x}=\dfrac{x}{\sqrt x-3}$
Vậy $C=\dfrac{x}{\sqrt x-3}$ (Đpcm).
