Lời giải:
a) Điều kiện: \(x\geq 0; x\neq 1\)
\(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
\(=\left ( \frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}(\sqrt{x}-1)}{(x+\sqrt{x}+1)(\sqrt{x}-1)}-\frac{x+\sqrt{x}+1}{(x+\sqrt{x}+1)(\sqrt{x}-1)}\right ):\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}:\frac{\sqrt{x}-1}{2}\)
\(=\frac{x-2\sqrt{x}+1}{x\sqrt{x}-1}:\frac{\sqrt{x}-1}{2}=\frac{(\sqrt{x}-1)^2}{x\sqrt{x}-1}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
b)
Ta thấy vì \(x\geq 0 \Rightarrow x+\sqrt{x}+1\geq 1\)
\(\Rightarrow A=\frac{2}{x+\sqrt{x}+1}\leq \frac{2}{1}=2\)
Vậy GTLN của A là 2 khi \(x=0\)
