\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=\left|x-2\right|+\left|x-4\right|+\left|x-3\right|\)
\(=\left|x-2\right|+\left|4-x\right|+\left|x-3\right|\ge\left|x-2+4-x\right|+\left|x-3\right|=2+ \left|x-3\right|\ge2\)Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x=3\\x-4\le0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x=3\\x\le4\end{matrix}\right.\Leftrightarrow x=3\)
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