`ĐKXĐ:x\ge0,x\ne9`

`1,x=4(TM)`

Thay `x=4` vào `B` có:

`B={1}/{\sqrt{4}-3}={1}/{2-3}={1}/{-1}=-1`

Vậy với `x=4` thì `B=-1`

`2,S=A-B`

`={x+3}/{x-9}+{2}/{\sqrt{x}+3}-{1}/{\sqrt{x}-3}` 

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

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

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

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

`={x+\sqrt{x}-6}/{(\sqrt{x}+3)(\sqrt{x}-3)}` 

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

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

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

`={\sqrt{x}-2}/{\sqrt{x}-3}` 

Vậy với `x\ge0,x\ne9` thì `S={\sqrt{x}-2}/{\sqrt{x}-3}` 

`3,S={\sqrt{x}+1}/{\sqrt{x}+2}` 

`⇔{\sqrt{x}-2}/{\sqrt{x}-3}={\sqrt{x}+1}/{\sqrt{x}+2}`

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

`⇒(\sqrt{x}-2)(\sqrt{x}+2)=(\sqrt{x}+1)(\sqrt{x}-3)` 

`⇔x-4=x-3\sqrt{x}+\sqrt{x}-3` 

`⇔x-4-x+3\sqrt{x}-\sqrt{x}+3=0` 

`⇔2\sqrt{x}-1=0` 

`⇔2\sqrt{x}=1` 

`⇔\sqrt{x}=1/2` 

`⇔x=1/4(TM)` 

Vậy với `x=1/4` thì `S={\sqrt{x}+1}/{\sqrt{x}+2}`