C=(x-1)(x+6)*(x+2)(x+3)
=(x^2+5x-6)(x^2+5x+6)
=(x^2+5x)^2-36>=-36
Dấu = xảy ra khi x^2+5x=0
=>x=0 hoặc x=-5
\(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(C=\left(x^2+6x-x-6\right)\left(x^2+2x+3x+6\right)\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-6^2\)
\(C=\left(x^2+5x\right)^2-36\)
Mà: \(\left(x^2+5x\right)^2\ge0\) nên \(C=\left(x^2+5x\right)-36\ge-36\)
Dấu "=" xảy ra:
\(\left(x^2+5x\right)-36=-36\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy: \(C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)