Ta có :
\(A=\left|x\right|+\left|8-x\right|\ge\left|x+\left(8-x\right)\right|\)
\(\Leftrightarrow A\ge8\)
Dấu "=" xảy ra khi :
\(x\left(8-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le0\\8-x\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\8-x\ge0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le0\\8\le x\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\8\ge x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\8\le x\le0\end{matrix}\right.\)
Vậy ..
Áp dụng tính chất: |A|+|B|≥|A+B|
Dấu "=" xảy ra⇔A.B≥0
Ta có: A=|x|+|8-x| ≥ |x+8-x|
⇒A≥8, ∀x
Dấu "=" xảy ra ⇔x.(8-x)≥0
⇔0≤x≤8
Vậy: Min(A)=8⇔0≤x≤8
