Ta có : \(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)\)
\(=\left(x^2+x\right)\left(x^2-3x+4x-12\right)\)
\(=\left(x^2+x\right)\left(x^2+x-12\right)\left(1\right)\)
Đặt \(x^2+x=t\)
\(\Rightarrow\left(1\right)\Leftrightarrow t\left(t-12\right)=t^2-12t=t^2-12t+36-36=\left(t-6\right)^2-36\)
Vì : \(\left(t-6\right)^2\ge0\)
\(\Rightarrow\left(t-6\right)^2-36\ge-36\)
Dấu " = " xảy ra khi \(t-6=0\)
\(t=0+6\)
\(t=6\)
\(\Rightarrow x^2+x+6\) \(x=2\) hoăc \(x=-3\)
Vậy \(MIN_B=-36\) khi \(x=2;x=-3\)
Ta có : \(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)=\left[x\left(x+1\right)\right].\left[\left(x-3\right)\left(x+4\right)\right]\)
\(=\left(x^2+x\right)\left(x^2+x-12\right)\)
Đặt \(t=x^2+x-6\) \(\Rightarrow B=\left(t+6\right)\left(t-6\right)=t^2-36\ge-36\)
Dấu "=" xảy ra khi \(t=0\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
Vậy Min B = -36 <=> \(\left[\begin{array}{nghiempt}x=-3\\x=2\end{array}\right.\)
