\(A=x\left(x+1\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt \(x^2+x=k\)
Lúc đó \(A=k\left(k-4\right)\)
\(=k^2-4k+4-4=\left(k-2\right)^2-4\ge-4\)
(Dấu "=" xảy ra khi \(k=2\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x^2+x-2=0\)
Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\))