Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A\ge\left|x+2+1-x\right|=\left|3\right|=3\)
Dấu " = " khi \(\left\{\begin{matrix}x+2\ge0\\1-x\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\ge-2\\x\le1\end{matrix}\right.\Rightarrow-2\le x\le1\)
\(\Rightarrow x\in\left\{-2;-1;0;1\right\}\)
Vậy \(MIN_A=3\) khi \(x\in\left\{-2;-1;0;1\right\}\)