1. `A=\sqrt{27}-\sqrt{12}-\sqrt{3}+3`
`=3\sqrt{3}-2\sqrt{3}-\sqrt{3}+3`
`=3`
Vậy `A=3`
`B=(x+1-2\sqrt{x})/(\sqrt{x}-1)+(x+\sqrt{x})/(\sqrt{x}+1)`
`=(\sqrt{x}-1)^2/(\sqrt{x}-1)+(\sqrt{x}(\sqrt{x}+1))/(\sqrt{x}+1)`
`=\sqrt{x}-1+\sqrt{x}`
`=2\sqrt{x}-1`
Vậy `B=2\sqrt{x}-1` với $x≥0,x\neq1$
2. Để `B=A⇔2\sqrt{x}-1=3`
`⇔2\sqrt{x}=4`
`⇔\sqrt{x}=2`
`⇔x=4` $(tm$ $đk:x≥0,x\neq1)$
Vậy `x=4`.
