`#huy`
`\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}`
`=>\sqrt{3}.x+\sqrt{3}=\sqrt{3}(x+1)`
`=>\sqrt{12}+\sqrt{27}=5+\sqrt{3}`
`=>\sqrt{3}(x+1)=5+\sqrt{3}`
`=>\sqrt{3}.x+\sqrt{3}=\sqrt{3^3}+2\sqrt{3}`
`=>\sqrt{3}-\sqrt{3^3}-\sqrt{3}=0`
`=>\sqrt{3}(x-4)=0`
`=>4`
Vậy `x=4`