Question

Tìm GTNN của biểu thức sau:

D=(x-1)(x+2)(x+3)(x+6)

Answer 1

\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-6x^2=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\Rightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\ge-36\)

\("="\Leftrightarrow\left(x^2+5x\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)