Lời giải:
ĐK: \(x\geq 0; x\neq 1\)
Ta có:
\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(=\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
\(=\frac{x+2+\sqrt{x}(\sqrt{x}-1)-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
\(=\frac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
Do đó:
\(P=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}:\frac{\sqrt{x}-1}{2}=\frac{2}{x+\sqrt{x}+1}\)
