ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{1}{\left(\sqrt{x}+1\right)}:\left(\frac{x-9-x+4+\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)=\frac{1}{\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=1-\frac{4}{\sqrt{x}+1}\)
Để P nguyên \(\Rightarrow\sqrt{x}+1=Ư\left(4\right)\)
Mà \(\sqrt{x}+1\ge1\Rightarrow\sqrt{x}+1=\left\{1;2;4\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3\right\}\Rightarrow x=\left\{0;1;9\right\}\)
Do \(x=9\) ko thuộc TXĐ \(\Rightarrow x=\left\{0;1\right\}\)
