`1, A=(\sqrt{x}-1)/(\sqrt{x}-3)` (`x>=0; x\ne 9`)
Thay `x=36(\text{tmđk})` vào A ta có:
`\qquad A=(sqrt{36}-1)/(\sqrt{36}-3)=(6-1)/(6-3)=5/3`
Vậy `A=5/3` khi `x=36`
`2,` Với `x>=0; x\ne 1` thì
`B=(\sqrt{x}+3)/(\sqrt{x}+1)-5/(1-\sqrt{x})+4/(x-1)`
`B=((\sqrt{x}+3)(\sqrt{x}-1)+5(\sqrt{x}+1)+4)/((\sqrt{x}-1)(\sqrt{x}+1))`
`B=(x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4)/((\sqrt{x}-1)(\sqrt{x}+1))`
`B=(x+7\sqrt{x}+6)/((\sqrt{x}-1)(\sqrt{x}+1))`
`B=(x+\sqrt{x}+6\sqrt{x}+6)/((\sqrt{x}-1)(\sqrt{x}+1))`
`B=((\sqrt{x}+1)(\sqrt{x}+6))/((\sqrt{x}-1)(\sqrt{x}+1))`
`B=(\sqrt{x}+6)/(\sqrt{x}-1)` (đpcm)
Vậy `B=(\sqrt{x}+6)/(\sqrt{x}-1)` với `x>=0; x\ne1`
`3,` Với `x>=0; x\ne1; x\ne9`
`\qquad A.B=3`
`<=> (\sqrt{x}-1)/(\sqrt{x}-3).(\sqrt{x}+6)/(\sqrt{x}-1)=3`
`<=> (\sqrt{x}+6)/(\sqrt{x}-3)-3=0`
`<=> (\sqrt{x}+6-3(\sqrt{x}-3))/(\sqrt{x}-3)=0`
`=> \sqrt{x}+6-3\sqrt{x}+9=0`
`<=> -2\sqrt{x}+15=0`
`<=> 2\sqrt{x}=15`
`<=> \sqrt{x}=15/2`
`<=> x=225/4 (\text{tm})`
Vậy `x=225/4` thì `A.B=3`
