\(1,P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}}\right)\\ =\left(\dfrac{x-1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{x-1}{\sqrt{x}}\right):\left(\dfrac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{x-1}{\sqrt{x}}\right)\times\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\\ =\dfrac{x-1}{1}\times\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(2,\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\dfrac{9}{2}\\ =>2\left(\sqrt{x}+1\right)^2=9\sqrt{x}\\ =>2x+4\sqrt{x}+2-9\sqrt{x}=0\\ =>2x-5\sqrt{x}+2=0\\ =>2x-4\sqrt{x}-\sqrt{x}+2=0\\ =>2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\\ =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\left(thoanman\right)\\x=\dfrac{1}{4}\left(thoaman\right)\end{matrix}\right.\)
Vậy \(x=4;x=\dfrac{1}{4}\)
