Question

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

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)