Question

  RÚT GọN(\(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\)):(\(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\) ) 

Answer 1

\(\left(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right):\left(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\right)\) (ĐK: \(x>0;x\ne4\))

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

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

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

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

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

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