Với mọi x ta có :
\(\left|x+7\right|=\left|-x-7\right|\)
\(\Leftrightarrow\left|x-3\right|+\left|x+7\right|=\left|x-3\right|+\left|-x-7\right|\)
\(\Leftrightarrow\left|x-3\right|+\left|-x-7\right|\ge\left|\left(x-3\right)+\left(-x-7\right)\right|\)
\(\Leftrightarrow\left|x-3\right|+\left|-x-7\right|\ge10\)
Mà \(\left|x+1\right|\ge0\)
\(\Leftrightarrow\left|x-3\right|+\left|-x-7\right|+\left|x+1\right|\ge10\)
\(\Leftrightarrow A\ge10\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left(x-3\right)\left(-x-7\right)\ge0\left(1\right)\\\left|x+1\right|=0\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\-x-7\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\-x-7\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\-7\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\-7\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\-7\le x\le3\left(I\right)\end{matrix}\right.\)
Từ \(\left(2\right)\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\left(II\right)\)
Từ \(\left(I\right)+\left(II\right)\Leftrightarrow A_{Min}=10\Leftrightarrow x=-1\)
Ta có A = |x - 3| + |x + 7| + |x + 1|
= |3 - x| + |x + 7| + |x +7|
≥ |3 - x + x + 1| + |x + 7| ≥ 4
Vậy MINA = 4 tại x = -7
