\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(P=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(P=\left(x^2+6x-x+6\right)\left(x^2+3x+2x+6\right)\)
\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(P=\left(x^2+5x\right)^2-6^2\)
Vì \(\left(x^2+5x\right)^2\ge0\Rightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\ge-36\)Dấu bằng xảy ra khi \(\left(x^2+5x\right)^2=0\)
\(\Leftrightarrow\) x= 0 hoặc x = -5
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\rightarrow min_A=-36\)
\(\leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
