Question

\((\dfrac{x}{\sqrt{x}-2}-\dfrac{x-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+6}{x-4}):(\dfrac{\sqrt{x}+2}{\sqrt{x}+2}-1)\)

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Answer 1

 

Sửa đề: \(\left(\dfrac{x}{\sqrt{x}-2}-\dfrac{x-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+6}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1\right)\)\(=\left(\dfrac{x\left(\sqrt{x}+2\right)-\left(x-1\right)\left(\sqrt{x}-2\right)-\sqrt{x}-6}{x-4}\right):\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\sqrt{x}-2}\)

\(=\dfrac{x\sqrt{x}+2x-x\sqrt{x}+2x+\sqrt{x}-2-\sqrt{x}-6}{x-4}:\dfrac{4}{\sqrt{x}-2}\)

\(=\dfrac{4x-8}{x-4}\cdot\dfrac{\sqrt{x}-2}{4}=\dfrac{x-2}{\sqrt{x}+2}\)