Question

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

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Answer 1

ĐKXĐ: \(x>0;x\ne1,4\)

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

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

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

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

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