`B=(3-5\sqrt{x})/(x-9)+(\sqrt{x}-1)/(\sqrt{x}-3) (x\ne9;x>0)`

`B=(3-5\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))+(\sqrt{x}-1)/(\sqrt{x}-3)`

`B=(3-5\sqrt{x}+(\sqrt{x}-1)(\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}+3))`

`B=(3-5\sqrt{x}+x+2\sqrt{x}-3)/((\sqrt{x}-3)(\sqrt{x}+3))`

`B=(x-3\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))`

`B=(\sqrt{x}(\sqrt{x}-3))/((\sqrt{x}-3)(\sqrt{x}+3))`

`B=(\sqrt{x})/(\sqrt{x}+3)`

`A.B>=2`

`(x+7)/(\sqrt{x}) . (\sqrt{x})/(\sqrt{x}+3)>=2`

`(x+7)/(\sqrt{x}+3)-2>=0`

`(x+7-2(\sqrt{x}+3))/(\sqrt{x}+3)>=0`

`(x+7-2\sqrt{x}-6)/(\sqrt{x}+3)>=0`

`(x-2\sqrt{x}+1)/(\sqrt{x}+3)>=0`

`((\sqrt{x}-1)^2)/(\sqrt{x}+3)>=0`

Ta có: `\sqrt{x}>0=>\sqrt{x}+3>0` nên

`(\sqrt{x}-1)^2>=0`

Mà `(\sqrt{x}-1)^2>=0 AA x>0` 

Vậy `x>0;x\ne9`