ĐK: \(x\ge0,x\ne1\)
a) \(P=\frac{x+2}{x\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{\sqrt{x}-1}{x-1}=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{x+2+x-1-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
b) Để \(\left|P\right|=\frac{2}{3}\) thì \(\left|\frac{\sqrt{x}}{x-\sqrt{x}+1}\right|=\frac{2}{3}\)(1)
Vì \(\sqrt{x}\ge0,x-\sqrt{x}+1\ge0\Rightarrow P\ge0\)
Vậy (1)\(\Leftrightarrow\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{2}{3}\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\Leftrightarrow2x-5\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy x=4 hoặc \(x=\frac{1}{4}\) thì \(\left|P\right|=\frac{2}{3}\)
c) Ta có \(\sqrt{x}\ne1\Rightarrow\left(\sqrt{x}-1\right)^2>0\Leftrightarrow x-2\sqrt{x}+1>0\Leftrightarrow x-\sqrt{x}+1>\sqrt{x}\Leftrightarrow\frac{\sqrt{x}}{x-\sqrt{x}+1}< 1\Leftrightarrow P< 1\)
Vậy với mọi giá trị của x làm cho P xác định thì P<1
