Lời giải:
ĐKXĐ: \(x> 0; x\neq 1\)
Có: \(P=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(P=\frac{3x+3\sqrt{x}-3}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(P=\frac{3x+3\sqrt{x}-3-(x-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(P=\frac{2x+3\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{(2\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(P=\frac{2\sqrt{x}-1}{\sqrt{x}-1}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}-\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(P=\frac{(2\sqrt{x}-1)\sqrt{x}-(\sqrt{x}-2)-(\sqrt{x}-2)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}\)
\(P=\frac{x+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Để \(P\in\mathbb{Z}\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}\in\mathbb{Z}\Leftrightarrow 1+\frac{2}{\sqrt{x}-1}\in\mathbb{Z}\Leftrightarrow \frac{2}{\sqrt{x}-1}\in\mathbb{Z}\)
\(\Leftrightarrow 2\vdots \sqrt{x}-1\Rightarrow \sqrt{x}-1\in\left\{\pm 1; \pm 2\right\}\)
\(\Rightarrow \sqrt{x}\in\left\{0; 2; 3\right\}\)
\(\Rightarrow x\in\left\{0; 4;9\right\}\)
Vì $x\neq 0$ nên \(x\in \left\{4; 9\right\}\)
c)
\(P=\sqrt{x}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\)
\(\Rightarrow \sqrt{x}+1=x-\sqrt{x}\Leftrightarrow x-2\sqrt{x}-1=0\)
\(\Rightarrow x=3+2\sqrt{2}\)
