ĐKXĐ: \(x>0;x\ne4;9\)
\(P=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{x+2\sqrt{x}+1-\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\right)\)
\(=\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right).\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P< 0\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}< 0\Rightarrow\sqrt{x}-2< 0\Rightarrow x< 4\)
Vậy để \(P< 0\Rightarrow0< x< 4\)
