Question

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

Answer 1

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

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

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

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

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