$A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}$ Rút gọn Tìm tc giâ trị x để A>1

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1. ĐKXĐ: $x≥0$ và $x\neq4$

   a) $A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt[]{x}-2}+\dfrac{1}{\sqrt[]{x}+2}$

   $=\dfrac{x+\sqrt[]{x}+2+\sqrt[]{x}-2}{(\sqrt[]{x}+2)(\sqrt[]{x}-2)}$

   $=\dfrac{\sqrt[]{x}(\sqrt[]{x}+2)}{(\sqrt[]{x}+2)(\sqrt[]{x}-2)}$

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

   b) $A>1$

   $↔ \dfrac{\sqrt[]{x}}{\sqrt[]{x}-2}>1$

   $↔ \dfrac{\sqrt[]{x}}{\sqrt[]{x}-2}-1>0$

   $↔ \dfrac{2}{\sqrt[]{x}-2}>0$

   $↔ \sqrt[]{x}-2>0$

   $↔ \sqrt[]{x}>2$

   $→ x>4$

   Trả lời

2. `A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}` `(x ≥ 0; x ne 4)`

   `=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}`

   `=\frac{x+2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}`

   `=\frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-2)}`

   `=\frac{\sqrt{x}}{\sqrt{x}-2}`

   $\text{Để A > 1}$

   `⇔\frac{\sqrt{x}}{\sqrt{x}-2}>1`

   `⇔\frac{\sqrt{x}}{\sqrt{x}-2}-1>0`

   `⇔\frac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0`

   `⇔\frac{2}{\sqrt{x}-2}>0`

   $\text{Có: 2 > 0}$

   `⇒\sqrt{x}-2>0`

   `⇔\sqrt{x}>2`

   `⇔x>4`

   $\text{kết hợp đkxđ}$

   $\text{⇒ Để A > 1 thì x > 4}$

   Trả lời