a, Ta có: \(A=\left|x-1\right|+\left|x-2017\right|=\left|x-1\right|+\left|2017-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A\ge\left|x-1+2017-x\right|=\left|-2016\right|=2016\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\2017-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le2017\end{matrix}\right.\Rightarrow1\le x\le2017\)
Vậy \(MIN_A=2016\) khi \(1\le x\le2017\)
b, Ta có: \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left|x-5\right|\ge0\end{matrix}\right.\Rightarrow\left(x-5\right)^2+\left|x-5\right|\ge0\)
\(\Rightarrow B=\left(x-5\right)^2+\left|x-5\right|+2014\ge2014\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-5\right)^2=0\\\left|x-5\right|=0\end{matrix}\right.\Rightarrow x=5\)
Vậy \(MIN_B=2014\) khi x = 5
