\(M=2018+\left(x-2019\right)^{2018}\)
Vì \(\left(x-2019\right)^{2018}\ge0\Rightarrow M\ge2018\)
Vậy Mmin = 2018 khi x = 2019
\(M=2018+\left(x-2019\right)^{2018}\)
Ta có : \(\left(x-2019\right)^{2018}\ge0\forall x\)
\(2018+\left(x-2019\right)^{2018}\ge2018\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2019\right)^{2018}=0\)
\(\Leftrightarrow x-2019=0\)
\(\Leftrightarrow x=2019\)
Vậy : min\(M=2018\) tại x = 2019.