Ta có
\(A=\left|x-1\right|+\left|x+2\right|=\left|1-x\right|+\left|x+2\right|\)
Ta có : \(\begin{cases}\left|1-x\right|\ge1-x\\\left|x+2\right|\ge x+2\end{cases}\)\(\Rightarrow\left|1-x\right|+\left|x+2\right|\ge1-x+x+2=3\)
Dấu " = " xảy ra khi \(\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\le1\\x\ge-2\end{cases}\)
Vậy MINA=3 khi \(-2\le x\le1\)