Vì \(\left|x-1\right|+\left|x-2\right|=1\)
\(\Rightarrow\left|x-1\right|+\left|2-x\right|=1\)
Ta có: \(\left[\begin{matrix}\left|x-1\right|\ge x-1\forall x\\\left|2-x\right|\ge2-x\forall x\end{matrix}\right.\)
Sử dụng \(BĐT:\)
\(\Rightarrow\left|x-1\right|+\left|2-x\right|\ge x-1+2-x\)
\(\Rightarrow\left|x-1\right|+\left|2-x\right|\ge1\)
Dấu \("="\) xảy ra khi \(\left[\begin{matrix}x-1\ge0\\2-x\ge0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\) \(\Rightarrow1\le x\le2\)
Vậy \(1\le x\le2.\)