Question

Tìm GTNN của:

A=(x-1)(x+2)(x+3)(x+6)+7

β

Answer 1

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+7\\ =\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+7\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)+7\\ =\left(x^2+5x\right)^2-6^2+7\\ =\left(x^2+5x\right)^2-36+7\\ =\left(x^2+5x\right)^2-29\ge-29\)

Dấu "=" xảy ra `<=> x^2 +5x= 0`

`<=>x(x+5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy GTNN của biểu thức là `-29` \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)