\(A=\left|x+2\right|+\left|x-1\right|\)
\(A=\left|x+2\right|+\left|1-x\right|\)
Áp dụng bđt: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(A=\left|x+2\right|+\left|1-x\right|\ge \left|x+2+1-x\right|=3\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2\ge0\Rightarrow x\ge-2\\1-x\ge0\Rightarrow x\le1\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\Rightarrow x< -2\\1-x< 0\Rightarrow x>1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(-2\le x\le1\)
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