Có: \(A>\frac{2016}{2016}+\frac{2017}{2017}=2\)
Có: \(B=\frac{4035}{4033}< 2\)
\(\Rightarrow A>B.\)
\(B=\frac{2017+2018}{2016+2017}=\frac{2017}{2016+2017}+\frac{2018}{2016+2017}\)
Ta có
\(\frac{2017}{2016+2017}< \frac{2017}{2016}\) ;
\(\frac{2018}{2017}< \frac{2018}{2017}\)
\(\Rightarrow B< \frac{2017}{2016}+\frac{2018}{2017}=A\)
Vậy B<A