Đáp án:
1) Thay `x=9` vào A ta được:
`A=\frac{\sqrt{9}}{\sqrt{9}+2}`
`= \frac{3}{3+2}`
`= 3/5`
2) `B=\frac{x}{x-4} +\frac{1}{\sqrt{x}-2} +\frac{1}{\sqrt{x}+2}`
`= \frac{x}{(\sqrt{x}-2)(\sqrt{x}+2)} +\frac{1}{\sqrt{x}-2} +\frac{1}{\sqrt{x}+2}`
`= \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}`
3) `A/B> -1`
`=> \frac{\sqrt{x}}{\sqrt{x}+2} : \frac{\sqrt{x}}{\sqrt{x}-2} > -1`
$\text{ĐK: 0<x<4, x>4}$
`=> \frac{\sqrt{x}-2}{ \sqrt{x}+2}+1>0`
`=> \frac{\sqrt{x}-2 +\sqrt{x}+2}{\sqrt{x}+2}>0`
`=> \frac{2\sqrt{x}}{\sqrt{x}+2}>0`
`=> 2\sqrt{x}>0`
`=> 2\sqrt{x}=0`
`=> x=0`
Vậy `0<x<4; x>4`
