Đáp án: $x>9$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0,x\ne 9$
Ta có:
$A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}$
$\to A=\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{(\sqrt{x}+1)(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3-11\sqrt{x}}{x-9}$
$\to A=\dfrac{2x-3\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{x+4\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3-11\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to A=\dfrac{2x-3\sqrt{x}+x+4\sqrt{x}+3-(3-11\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to A=\dfrac{3x+12\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}$
Để $A\ge0$
$\to \dfrac{3x+12\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}\ge 0$
$\to \sqrt{x}-3\ge 0$ vì $x\ge 0$
$\to \sqrt x\ge 3$
$\to x\ge 9$ mà $x\ne 9$
$\to x>9$
