Ta có:
\(A=\frac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}=\frac{\left|x-2016\right|+2018-1}{\left|x-2016\right|+2018}=1-\frac{1}{\left|x-2016\right|+2018}\)
Vì \(\left|x-2016\right|\ge0\Rightarrow\left|x-2016\right|+2018\ge2018\Rightarrow\frac{1}{\left|x-2016\right|+2018}\le\frac{1}{2018}\)
=>\(A=1-\frac{1}{\left|x-2016\right|+2018}\ge\frac{2017}{2018}\)
=>\(A_{min}=\frac{2017}{2018}\)<=>|x-2016|=0<=>x-2016=0<=>x=2016