a) \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\left(đk:x>0\right)\)
\(=\dfrac{x-1+\sqrt{x}\left(x-\sqrt{x}\right)}{\left(x-1\right)\left(x-\sqrt{x}\right)}.\dfrac{x\sqrt{x}-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{x\sqrt{x}-1}{\left(x-1\right)\left(x-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(x-1\right)}{x\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x-1\right)}{\sqrt{x}\left(x-1\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-1}\)
b) \(P=\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
a: Ta có: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
b: Để \(P=\dfrac{1}{2}\) thì \(\sqrt{x}-1=2\)
hay x=9
