\(\left|x-3\right|+\left|x-5\right|=\left|x-3\right|+\left|5-x\right|\)
Áp dụng bđt:
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|=2\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\Rightarrow x\ge3\\5-x\ge0\Rightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\5-x< 0\Rightarrow x>5\end{matrix}\right.\end{matrix}\right.\)
Vậy \(3\le x\le5\)
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