\(C=\left|x+7\right|+\left|x-5\right|+\left|x-1\right|=\left(\left|x+7\right|+\left|5-x\right|\right)+\left|x-1\right|\)
Ta có: \(\left|x+7\right|+\left|5-x\right|\ge\left|x+7+5-x\right|=8\)
Mà \(\left|x-1\right|\ge0\)
\(\Rightarrow C=\left(\left|x+7\right|+\left|5-x\right|\right)+\left|x-1\right|\ge12+0=12\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+7\right)\left(5-x\right)\ge0\\\left|x-1\right|=0\end{cases}\Rightarrow\hept{\begin{cases}-7\le x\le5\\x=1\end{cases}}\Rightarrow x=1}\)
Vậy Cmin = 12 khi x = 1