Ý 1:
\(P\left(x\right)=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{x}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
Ý 2:
\(P\left(x\right)=\frac{2\sqrt{x}}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}\)
Vì: \(\sqrt{x}+\frac{1}{\sqrt{x}}-1>1\) nên:
\(\Rightarrow\sqrt{x}+\frac{1}{\sqrt{x}}-1=2\)
Để \(Q\) nguyên \(\Leftrightarrow\sqrt{x}+\frac{1}{\sqrt{x}}-1\inƯ\left(2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\frac{7+3\sqrt{5}}{2}\\x_2=\frac{7-3\sqrt{5}}{2}\end{matrix}\right.\)
Vậy .................................
