\(A=\left|x-2006\right|+\left|x-1\right|=\left|x-2006\right|+\left|-x+1\right|\ge\left|x-2006-x+1\right|=2005\)
dấu = xảy ra khi \(\left(x-2006\right).\left(-x+1\right)\ge0\)
\(\Rightarrow1\le x\le2006\)
Vậy Min A=2015 khi và chỉ khi \(1\le x\le2006\)
\(A=|x-2006|+|x-1|=|x-2006|+|1-x|\)
\(\Rightarrow A\ge|x-2006+1-x|=|-2005|=2005\)
\(\Rightarrow minA=2005\Leftrightarrow\left(x-2006\right).\left(1-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-2006< 0\\1-x< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\1< x\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\x>1\end{cases}}\Rightarrow1< x< 2006\left(t/m\right)\)
\(TH2:\hept{\begin{cases}x-2006\ge0\\1-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\1\ge x\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\x\le1\end{cases}}\)(vô lý)
Vậy \(minA=2005\Leftrightarrow1< x< 2006\)
Áp dụng BĐT sau : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Ta có :
\(A=\left|x-2006\right|+\left|x-1\right|=\left|x-2006\right|+\left|1-x\right|\ge\left|x-2006+1-x\right|\)
\(\Rightarrow A\ge2005\)
Dấu''='' xảy ra \(\Leftrightarrow\left(x-2006\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2006\ge0\\1-x\ge0\end{cases}}\)hoặc \(\orbr{\begin{cases}x-2006< 0\\1-x< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge2006\\x\le1\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2006\\x>1\end{cases}}\)(loại )
\(\Leftrightarrow1\le x\le2006\)
Vậy ..................
