\(B=\left|x-2016\right|+\left|x-2017\right|\)
\(B=\left|x-2016\right|+\left|x-2017\right|\)
\(B=\left|x-2016\right|+\left|2017-x\right|\)
Áp dụng bđt:
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(B\ge\left|x-2016+2017-x\right|=1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2016\ge0\Rightarrow x\ge2016\\2017-x\ge0\Rightarrow x\le2017\end{matrix}\right.\\\left\{{}\begin{matrix}x-2016< 0\Rightarrow x< 2016\\2017-x< 0\Rightarrow x>2017\end{matrix}\right.\end{matrix}\right.\)
Vậy \(2016\le x\le2017\)
Ta có :
\(B=\left|x-2016\right|+\left|x-2017\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có :
\(B=\left|x-2016\right|+\left|x-2017\right|=\left|x-2016\right|+\left|2017-x\right|\ge\left|x-2016+2017-x\right|=1\)
Dấu "=" xảy ra khi :
\(\left(x-2016\right)\left(2017-x\right)\ge0\)
\(\Leftrightarrow2016\le x\le2017\)
Vậy GTNN của B = 1 khi \(2016\le x\le2017\)
Ta có:
\(B=\left|x-2016\right|+\left|x-2017\right|\)
\(=\left|x-2016\right|+\left|2017-x\right|\ge\left|x-2016+2017-x\right|=1\)Khi \(\left(x-2016\right)\left(2017-x\right)\ge0\)
\(\Rightarrow2016\le x\le2017\)
