Question

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}\). \(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
Giúp em giải chi tiết các bước rút gọn cho pt này với
 

Answer 1

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

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

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

Answer 2

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2}\) (\(đk:x\ge0;x\ne\sqrt{2}\))

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

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