a) điều kiện xác định : \(x\ge0;x\ne1\)
ta có : \(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
b) để \(P=\dfrac{2}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(\sqrt{x}-1\right)=2\left(x+\sqrt{x}+1\right)\Leftrightarrow3\sqrt{x}-3=2x+2\sqrt{x}+2\)
\(\Leftrightarrow2x-\sqrt{x}+5=0\Leftrightarrow2\left(x-\dfrac{1}{2}\sqrt{x}+\dfrac{1}{16}\right)+\dfrac{79}{16}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{4}\right)^2+\dfrac{79}{16}=0\left(vôlí\right)\)
vậy không tồn tại \(x\) để \(P=\dfrac{2}{3}\)
