\(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|x-2006+2007-x\right|=\left|1\right|=1\)

\(minA=1\Leftrightarrow\left(x-2006\right)\left(2007-x\right)\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2006\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2006\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow2006\le x\le2007\)

\(A=\left|x-2006\right|+\left|2007-x\right|\)

Vì \(x>2007\) nên \(2x-4013>4014-4013=1\)

\(\Rightarrow A>1\)

Vậy \(A_{min}=1\Leftrightarrow2006\le x\le2007\)

Ta có :

\(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|x-2006+2007-x\right|\)

\(\Rightarrow A\ge1\)

\(\Rightarrow A_{min}=1\)

\(\Leftrightarrow\left(x-2006\right)\left(2007-x\right)\ge0\)

Ta có bảng xét dấu :

\(\Rightarrow2006\le x\le2007\)

2005<x<2008

Ta có :

A=|x−2006|+|2007−x|≥|x−2006+2007−x|

⇒A≥1

⇒Amin=1

⇔(x−2006)(2007−x)≥0

Ta có bảng xét dấu :

xx-2006( x - 2006 )( 2007 - x )20062007002007-x0__++++00+__

⇒2006≤x≤2007

Ta có : \(A=\left|x-2006\right|+\left|2007-x\right|\)

\(=\left|2006-x\right|+\left|x-2007\right|\)

Ta có : \(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|2006-x+x-2007\right|=1\)

Dấu " = " xảy ra khi và chỉ \(2006\le x\le2007\)

Vậy GTNN \(A=1\)khi \(2006\le x\le2007\)