Ta co: \(\left\{{}\begin{matrix}\left|x\right|\ge x\\\left|8-x\right|\ge8-x\end{matrix}\right.\:\Rightarrow A\ge x+8-x=8\Rightarrow A_{min}=8\)
Dâu "=" xay ra <=> x(8-x) \(\ge0\)
\(+,\left\{{}\begin{matrix}x\le0\\8-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\left(voli\right)\)
\(+,\left\{{}\begin{matrix}x\ge0\\8-x\ge0\end{matrix}\right.\Rightarrow0\le x\le8\)
Vậy:\(A_{min}=8\Leftrightarrow0\le x\le8\)
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