a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) Để \(P=\dfrac{3}{2}\) thì \(4\sqrt{x}+2=3\sqrt{x}+3\)
\(\Leftrightarrow x=1\)(Vô lý)