Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac1{1-\sqrt{x}}):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}$

$\to A=(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac1{\sqrt{x}-1}):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}$

$\to A=\dfrac{x+2+\sqrt{x}(\sqrt{x}-1)-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}:\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}$

$\to A=\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}$

$\to A=\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}$

$\to A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}$

b.Để $A\le 0$

$\to \dfrac{\sqrt{x}-1}{\sqrt{x}+2}\le 0$

$\to \sqrt{x}-1\le 0$ vì $\sqrt{x}+2>0$

$\to \sqrt{x}\le 1$

$\to 0\le x\le 1$

c.Ta có:

$A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}$

$\to A=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}$

$\to A=1-\dfrac3{\sqrt{x}+2}\ge 1-\dfrac3{0+2}=-\dfrac12$

Dấu = xảy ra khi $x=0$