Question

Answer 1

\(P=\left(\dfrac{1}{\sqrt{x}}-\dfrac{2}{1+\sqrt{x}}\right).\dfrac{x+\sqrt{x}}{1-\sqrt{x}}\left(đk:x>0,x\ne1\right)\)

\(=\dfrac{1+\sqrt{x}-2\sqrt{x}}{x+\sqrt{x}}.\dfrac{x+\sqrt{x}}{1-\sqrt{x}}=\dfrac{1-\sqrt{x}}{1-\sqrt{x}}=1\)

Answer 2

Ta có: \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}\right)\cdot\dfrac{x+\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1-\sqrt{x}}\)

\(=\dfrac{3\sqrt{x}+1}{1-\sqrt{x}}\)