ĐK:x>0,x≠0,x≠1
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{x-1}\right)=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{x-\sqrt{x}}\)b) Khi x=\(3+2\sqrt{2}\) thì \(P=\dfrac{\sqrt{3+2\sqrt{2}}-3}{3+2\sqrt{2}-\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}+1}-3}{3+2\sqrt{2}-\sqrt{2+2\sqrt{2}+1}}=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-3}{3+2\sqrt{2}-\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}+1-3}{3+2\sqrt{2}-\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{2+\sqrt{2}}=\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\)
c) Ta có \(P< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{x-\sqrt{x}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\x-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\x-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1< x< 9\)
Vậy 1<x<9 thì P<0