Lời giải:
a) ĐKXĐ: \(x\geq 0, x\neq 1\)
Ta có:
\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x(\sqrt{x}-1)+(\sqrt{x}-1)}\right):\left(\frac{x+\sqrt{x}}{x(\sqrt{x}+1)+(\sqrt{x}+1)}+\frac{1}{x+1}\right)\)
\(=\frac{x+1-2\sqrt{x}}{(x+1)(\sqrt{x}-1)}:\left(\frac{\sqrt{x}(\sqrt{x}+1)}{(x+1)(\sqrt{x}+1)}+\frac{1}{x+1}\right)\)
\(=\frac{(\sqrt{x}-1)^2}{(x+1)(\sqrt{x}-1)}:\left(\frac{\sqrt{x}}{x+1}+\frac{1}{x+1}\right)\)
\(=\frac{\sqrt{x}-1}{x+1}.\frac{x+1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\sqrt{x}-2\)
\(\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+1}=\sqrt{x}-2\)
\(\Rightarrow \sqrt{x}-1=(\sqrt{x}-2)(\sqrt{x}+1)=x-\sqrt{x}-2\)
\(\Rightarrow x-2\sqrt{x}-1=0\)
\(\Leftrightarrow (\sqrt{x}-1)^2=2\Rightarrow \left[\begin{matrix} \sqrt{x}-1=\sqrt{2}\rightarrow x=3+2\sqrt{2}\\ \sqrt{x}-1=-\sqrt{2}\rightarrow \sqrt{x}=1-\sqrt{2}< 0(\text{vô lý})\end{matrix}\right.\)
Vậy \(x=3+2\sqrt{2}\)
c)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Vì \(\sqrt{x}\geq 0\Rightarrow \frac{2}{\sqrt{x}+1}\leq \frac{2}{0+1}=2\)
\(\Rightarrow A=1-\frac{2}{\sqrt{x}+1}\geq 1-2=-1\)
Vậy $A$ min bằng $-1$. Dấu bằng xảy ra khi $x=0$
