\(A=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
\(minA=-56\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(B=4x-x^2+1=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)
\(maxB=5\Leftrightarrow x=2\)
MinA=0
⇔x=1 hoặc x=-3 hoặc x=-2 hặc x=-6
B\(=-x^2+2x+1+2x\)
\(=-\left(x^2-2x+1\right)+2\left(1+x\right)\)
\(=-\left(x-1\right)^2-2\left(x-1\right)\)