Giải câu 5 đề 7 ôn thi toán lớp 9 lên 10.
$A=\left ( \frac{\sqrt{x}}{3+\sqrt{x}}+\frac{a+9}{9-x} \right ):\left ( \frac{3\sqrt{x}+1}{x-3\sqrt{x}+\frac{2}{\sqrt{x}}} \right )(x>0,x\neq 9)$
$=\left ( \frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{(3+\sqrt{x})(3-\sqrt{x})} \right ):\left ( \frac{3\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-3)}+\frac{2}{\sqrt{x}} \right )$
$=\frac{\sqrt{x}(3-\sqrt{x})+x+9}{(3+\sqrt{x})}:\frac{3\sqrt{x}+1+2(\sqrt{x}-3)}{\sqrt{x}(\sqrt{x-3})}$
$=\frac{3(\sqrt{x}+3)}{(3+\sqrt{x})(3-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-3)}{5\sqrt{x}-5}$
$=\frac{3\sqrt{x}}{5-5\sqrt{x}}$
$A>0\Leftrightarrow \frac{3\sqrt{x}}{5-5\sqrt{x}}>0\Leftrightarrow 5-5\sqrt{x}>0\Leftrightarrow \sqrt{x}<1\Leftrightarrow x<1$
Vậy $A > 0$ khi $0 < x < 1$