ĐKXĐ : \(x>0\) và \(x\ne1\)
Câu a : \(P=\left(\dfrac{2-x}{x-\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2-x+\sqrt{x}+x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu b : Thay \(x=\dfrac{9}{16}\) vào P ta được :
\(P=\dfrac{\sqrt{\dfrac{9}{16}}-1}{\sqrt{\dfrac{9}{16}}}=\dfrac{\dfrac{3}{4}-1}{\dfrac{3}{4}}=\dfrac{\dfrac{-1}{4}}{\dfrac{3}{4}}=-\dfrac{1}{3}\)
Câu c : Để \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
