# Violympic toán 7

$\left|x+2016\right|+\left|x+2017\right|+\left|x+2018\right|$

tìm GTNN

30 tháng 3 2018 lúc 19:40

Với mọi x ta có :

$\left|x+2018\right|=\left|-x-2018\right|$

$\Leftrightarrow\left|x+2016\right|+\left|x+2018\right|=\left|x+2016\right|+\left|-x-2018\right|$

$\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge\left|\left(x+2016\right)+\left(-x-2018\right)\right|$

$\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge\left|-2\right|$

$\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge2$

$\left|x+2017\right|\ge0$

$\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|+\left|x+2017\right|\ge2$

Dấu "=" xảy ra khi :

$\left\{{}\begin{matrix}\left(x+2016\right)\left(-x-2018\right)\ge0\left(1\right)\\\left|x+2017\right|=0\left(2\right)\end{matrix}\right.$

Từ $\left(1\right)$ $\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2016\ge0\\-x-2018\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2016\le0\\-x-2018\le0\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-2016\\-2018\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-2016\\-2018\le x\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}-2016\ge x\ge-2018\\x\in\varnothing\end{matrix}\right.$

$\Leftrightarrow-2016\ge x\ge-2018\left(I\right)$

Từ $\left(2\right)\Leftrightarrow x+2017=0$

$\Leftrightarrow x=-2017\left(II\right)$

Từ $\left(I\right)+\left(II\right)\Leftrightarrow GTNN$ của $\left|x+2016\right|+\left|x+2017\right|+\left|x+2017\right|=2\Leftrightarrow x=-2017$

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