Cho N= căn x/ căn x+ 2 +4 căn x/ x-4 với x > hoặc bằng 0; x khác 4

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1. a) $N = \dfrac{\sqrt x}{\sqrt x + 2} + \dfrac{4 \sqrt x}{x -4} (Đk: x \geq 0; x \neq 4)$

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

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

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

   $=\dfrac{x + 2 \sqrt x}{(\sqrt x + 2)(\sqrt x – 2)}$

   $=\dfrac{\sqrt x(\sqrt x  + 2)}{(\sqrt x + 2)(\sqrt x – 2)}$

   $=\dfrac{\sqrt x}{\sqrt x -2 }$

   b) Khi $N = 2$ thì $\dfrac{\sqrt x }{\sqrt x – 2} = 2$

             ⇔ $\sqrt x  = 2(\sqrt x – 2)$

             ⇔ $\sqrt x = 2\sqrt x – 4$

             ⇔ $\sqrt x = 4$
             ⇒  $x = 16 (T/m)$

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2. a/ $\dfrac{\sqrt x}{\sqrt x+2}+\dfrac{4\sqrt x}{x-4}\\=\dfrac{\sqrt x(\sqrt x-2)}{(\sqrt x-2)(\sqrt x+2)}+\dfrac{4\sqrt x}{(\sqrt x-2)(\sqrt x+2)}\\=\dfrac{x-2\sqrt x+4\sqrt 2}{(\sqrt x-2)(\sqrt x+2)}\\=\dfrac{x+2\sqrt x}{(\sqrt x-2)(\sqrt x+2)}\\=\dfrac{\sqrt x(\sqrt x+2)}{(\sqrt x-2)(\sqrt x+2)}\\=\dfrac{\sqrt x}{\sqrt x-2}(x≥0;x\ne 0)$

   b/ $N=2\\→\dfrac{\sqrt x}{\sqrt x-2}=2\\↔\sqrt x=2\sqrt x-4\\↔\sqrt x-2\sqrt x=-4\\↔-\sqrt x=-4\\↔\sqrt x=4\\↔x=16(TM)$

   Vậy $N=2$ khi $x=16$

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