B=\(\dfrac{2016+2017}{2017+2018}\)=\(\dfrac{2016}{2017+2018}+\dfrac{2017}{2017+2018}\)
Ta có:
\(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)\(^{\left(1\right)}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)\(^{\left(2\right)}\)
Cộng vế với vế của biểu thức (1), (2) suy ra:
\(\dfrac{2016}{2017+2018}+\dfrac{2017}{2017+2018}< \dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Hay A