Đáp án: $0\le x<9$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0,x\ne 9$
Ta có:
$P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}$
$\to P=\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3x+9}{x-9}$
$\to P=\dfrac{2x-6\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{x+3\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-(3x+9)}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to P=\dfrac{-3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to P=\dfrac{-3(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$\to P=\dfrac{-3}{\sqrt{x}-3}$
Để $P>0$
$\to \dfrac{-3}{\sqrt{x}-3}>0$
$\to \sqrt{x}-3<0$
$\to\sqrt{x}<3$
$\to 0\le x<9$
