Question

Rút gọn biểu thức

\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\) 

Answer 1

\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\) (ĐK: x ≥ 0, x ≠ 4)

\(=\left[\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right]\)

\(=\left(\dfrac{-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{6}{\sqrt{x}+2}\)

\(=\dfrac{\left(-6\right)\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

Vậy...