a/ ĐKXĐ: \(x>0,x\ne1,x\ne2\)
b/
\(P=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right]:\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right]\)
= \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
c/ Với \(x>0,x\ne1,x\ne2\)
Để P=\(\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=8\)
\(\Leftrightarrow x=64\left(tm\right)\)
Vậy để \(P=\dfrac{1}{4}\) thì x=64
