Ta có:A =\(\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(\Rightarrow\) A< \(\dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\)=\(\dfrac{2017^{2017}+2017}{2017^{2018}+2017}=\dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}=\dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)Vậy A