\(3 \left|x\right|+\left|x-2\right|=2\)
\(\Rightarrow2\left|x\right|+\left|x\right|+\left|2-x\right|=2\)
Ta có: \(\left|x\right|+\left|2-x\right|+2\left|x\right|\ge\left|x+2-x\right|+2\left|x\right|=2+2\left|x\right|\ge2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}0\le x\le2\\x=0\end{matrix}\right.\Leftrightarrow x=0\)