\(\Leftrightarrow A=\left(x^2+x\right)\left(x^2+x-4\right)\) Đặt x^2+x-a có
\(A=a\left(a-4\right)=a^2-4a\ge-4\)
Min A=-4 với a-2=0\(\Rightarrow x^2+x=2\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{3}{2}\\x+\frac{1}{2}=-\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)