Đáp án:
$\dfrac{2}{x+\sqrt{x}+1}$
Giải thích các bước giải:
`(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}):\frac{\sqrt{x}-1}{2}` ĐK: $x\neq1;\,x\ge0$
$=\bigg(\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\bigg).\dfrac{2}{\sqrt{x}-1}$
$=\dfrac{x+2+\sqrt{x}(\sqrt{x}-1)-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\dfrac{2}{\sqrt{x}-1}$
$=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\dfrac{2}{\sqrt{x}-1}$
$=\dfrac{2(x-2\sqrt{x}+1)}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}$
$=\dfrac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}$
$=\dfrac{2}{x+\sqrt{x}+1}$
