A=|x-2006|+|2007-x|
\(\ge\)\(\)|x-2006+2007-x|
=|1|
=1
<=>A\(\ge\)1.Dấu bằng xảy ra khi(x-2006)(2007-x)\(\ge\)0
\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2006\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2006\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2006\\x\le2007\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2006\\x\ge2007\end{matrix}\right.\end{matrix}\right.\)
\(< =>2006\le x\le2007\)
Vậy Min A=1 tại\(2006\le x\le2007\)
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