\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1-1=\left(x^2+3x+1\right)^2-1\ge-1\)
Vậy MIN bt là -1 với \(x^2+3x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{matrix}\right.\)