\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " khi \(\left(x^2+5x\right)^2=0\Rightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy MIN P = -36 khi x = 0 hoặc x = -5
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow P=\left[\left(x+1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(\Leftrightarrow\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)\)
\(\Leftrightarrow P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(\Leftrightarrow P=\left(x^2+5x\right)^2-36\)
Ta có \(\left(x^2+5x\right)^2\ge0\) với mọi \(x\in Q\)
nên với \(P=\left(x^2+5x\right)^2-36\) thì \(P\) đạt giá trị nhỏ nhất khi \(\left(x^2+5x\right)^2=0\)
Lúc đó ta có \(x^2+5x=0\Rightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(MinP=-36\) đạt được khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(P=\) \(\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(P=\left(x^2+5x\right)^2-36\)
\(\left(x^2+5x\right)^2\ge0\)
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi:
\(x^2+5x=0\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
