Giải thích các bước giải:

1.ĐKXĐ: $x\ge 0, x\ne 1$

Ta có:

$A=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac2{\sqrt{x}+1}+\dfrac2{1-x}$

$\to A=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac2{\sqrt{x}+1}-\dfrac2{x-1}$

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

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

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

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

b.Để $A<\dfrac12$

$\to \dfrac{\sqrt{x}}{\sqrt{x}+1}<\dfrac12$

$\to 2\sqrt{x}<\sqrt{x}+1$

$\to \sqrt{x}<1$

$\to 0\le x<1$

3.Ta có

$A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\ge 0$

$\to GTNN_A=0\to x=0$