Vì \(2016^{2016}+1< 2016^{2017}+1\) nên \(\frac{2016^{2016}+1}{2016^{2017}+1}< 1\)
\(\Rightarrow A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2017}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)Vậy A < B