Question

Rút gọn B

B = \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x+2}{x-\sqrt{x}-2}\)               (\(x\ge0\), \(x\ne4\))

Answer 1

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