\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-36\)
Ta có: \(\left(x^2+5x\right)^2\ge0\forall x\)
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\forall x\)
\(C=-36\Leftrightarrow\left(x^2+5x\right)^2=0\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy \(C_{min}=-36\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
C = ( x - 1 )( x + 3 )( x + 2 )( x + 6 )
C = [( x - 1 )( x + 6 )][( x + 3 )( x + 2 )]
C = ( x2 + 5x - 6 )( x2 + 5x + 6 )
Đặt a = x2 + 5x
=> C = ( a - 6 )( a + 6 ) = a2 - 36
\(a^2\ge0\forall a\Rightarrow a^2-36\ge-36\)
Dấu " = " xảy ra <=> a2 = 0 => a = 0
<=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy CMin = -36, đạt được khi x = 0 hoặc x = -5
