Đáp án:
$A>B$
Giải thích các bước giải:
$A=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\\
B=\dfrac{2016+2017+2018}{2017+2018+2019}\\
=\dfrac{2016}{2017+2018+2019}+\dfrac{2017}{2017+2018+2019}+\dfrac{2018}{2017+2018+2019}$
Do $2017<2017+2018+2019\Rightarrow \dfrac{1}{2017}>\dfrac{1}{2017+2018+2019}$
$\Rightarrow \dfrac{2016}{2017}>\dfrac{2016}{2017+2018+2019}$
Do $2018<2017+2018+2019\Rightarrow \dfrac{1}{2018}>\dfrac{1}{2017+2018+2019}$
$\Rightarrow \dfrac{2017}{2018}>\dfrac{2017}{2017+2018+2019}$
Do $2019<2017+2018+2019 \Rightarrow \dfrac{1}{2019}> \dfrac{1}{2017+2018+2019}$
$\Rightarrow \dfrac{2018}{2019}>\dfrac{2018}{2017+2018+2019}\\
\Rightarrow \dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2016+2017+2018}{2017+2018+2019}\\ A>B$