Ta có: \(A=\left(x-2\right)\left(x-3\right)\left(x^2-5x-6\right)\)
\(=\left(x^2-5x+6\right)\left(x^2-5x-6\right)\)
\(=\left(x^2-5x\right)^2-36\)
Ta có: \(\left(x^2-5x\right)^2\ge0\forall x\)
\(\Rightarrow\left(x^2-5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi
\(\left(x^2-5x\right)^2=0\Leftrightarrow x^2-5x=0\)\(\Leftrightarrow x\left(x-5\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy: GTNN của đa thức \(A=\left(x-2\right)\left(x-3\right)\left(x^2-5x-6\right)\) là -36 khi x∈{0;5}
