Đặt \(A_1=\left|x-15\right|+\left|x-17\right|\)
\(\Rightarrow A_1\ge\left|x-15+17-x\right|\forall x\)
\(\Rightarrow A_1\ge2\forall x\left(1\right)\)
\(\left|x-16\right|\ge0\forall x\left(2\right)\)
Từ (1) ; (2)
\(\Rightarrow A=A_1+\left|x-16\right|\ge2+0=2\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-15\right)\left(x-17\right)\ge0\\\left|x-16\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-16\right|=0\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x-15\ge0\\17-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-15\le0\\17-x\le0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=16\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge15\\x\le17\end{matrix}\right.\\\left\{{}\begin{matrix}x\le15\\x\ge17\left(VL\right)\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=16\\15\le x\le17\end{matrix}\right.\)
\(\Leftrightarrow x=16\)
Vậy Min A là : \(2\Leftrightarrow x=16\)
