`# T e a m n g u v u o t t r o i`
`B=3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}(x>=0)`
`=3x-\sqrt{3^2 .3}+\sqrt{\frac{x^2(x+3)}[x+3}}`
`=3x-3\sqrt{3}+\sqrt{x^2}`
`=3x-3\sqrt{3}+|x|`
`=3x-3\sqrt{3}+x(x>=0)`
`=4x-3\sqrt{3}`
Tại `x=\sqrt{3}`
`=4\sqrt{3}-3\sqrt{3}`
`=\sqrt{3}`
Vậy tại `x=\sqrt{3}<=>B=\sqrt{3}`
