Đặt \(A=\left|x-2001\right|+\left|x-1\right|=\left|2011-x\right|+\left|x-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|2011-x\right|+\left|x-1\right|\ge\left|2011-x+x-1\right|=\left|2010\right|=2010\)
Dấu " = " khi \(\left\{{}\begin{matrix}2011-x\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2010\\x\ge1\end{matrix}\right.\Rightarrow1\le x\le2010\)
Vậy \(MIN_A=2010\) khi \(1\le x\le2001\)
Đặt \(A=\left|x-2001\right|+\left|x-1\right|\)
Ta có:
\(\left(\left|a\right|+\left|b\right|\right)^2\ge\left(\left|a+b\right|\right)^2\)
\(\Leftrightarrow a^2+b^2+2\left|ab\right|\ge a^2+b^2+2ab\)
\(\Leftrightarrow\left|ab\right|\ge ab\)
Dấu \("="\) xảy ra khi \(ab\ge0\)
Áp dụng vào bài ta có:
\(\left|x-2001\right|+\left|x-1\right|\ge\left|x-2001+1-x\right|=2000\)
Dấu \("="\) xảy ra \(\left(x-2001\right)\left(x-1\right)\ge0\Rightarrow1\le x\le2001\)
\(\Leftrightarrow\left\{\begin{matrix}\left(x-2001\right)\left(x-1\right)\\1\le x\le2001\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=1\\x=2001\end{matrix}\right.\)
\(Min_A=2000\) xảy ra khi \(\left\{\begin{matrix}x=1\\x=2001\end{matrix}\right.\)
Làm lại:
Đặt \(A=\left|x-2001\right|+\left|x-1\right|=\left|2001-x\right|+\left|x-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A\ge\left|2001-x+x-1\right|=\left|2000\right|=200\)
Dấu " = " khi \(\left\{{}\begin{matrix}2001-x\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2001\\x\ge1\end{matrix}\right.\Rightarrow1\le x\le2001\)
Vậy \(MIN_A=2000\) khi \(1\le x\le2001\)
Đặt A=\(\left|x-2001\right|+\left|x-1\right|=\left|2001-x\right|+\left|x-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A\ge\left|2001-x+x-1\right|=\left|2000\right|=2000\)
Dấu "=" xảy ra khi : \(\sum_{x-1\ge0}^{2001-x\ge0}\Rightarrow\sum_{x\ge1}^{x\le2001}\Rightarrow1\le x\le2001\)
Vậy GTNN của A= 2000 khi \(1\le x\le2001\)
