P=\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)(đk: \(x\ge0;x\ne1\))
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) \(-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) \(-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\left(x+\sqrt{x}+1\right)}\)
để \(P=\frac{2}{7}\) thì:
\(\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\) \(\Leftrightarrow2x+2\sqrt{x}+2=7\sqrt{x}\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\) \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\) (tm)
vậy x=4 hoặc x=1/4 thì P=2/7
