Giải thích các bước giải:
Ta có :
$B=\dfrac{2018}{1}+\dfrac{2017}{2}+..+\dfrac{2}{2017}+\dfrac{1}{2018}$
$\to B=1+(1+\dfrac{2017}{2})+..+(\dfrac{2}{2017}+1)+(\dfrac{1}{2018}+1)$
$\to B=1+\dfrac{2+2017}{2}+..+\dfrac{2017+2}{2017}+\dfrac{2018+1}{2018}$
$\to B=1+\dfrac{2019}{2}+..+\dfrac{2019}{2017}+\dfrac{2019}{2018}$
$\to B=\dfrac{2019}{2}+..+\dfrac{2019}{2017}+\dfrac{2019}{2018}+1$
$\to B=2019(\dfrac{1}{2}+..+\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019})$
$\to B=2019A$
$\to \dfrac{A}{B}=\dfrac{1}{2019}<\dfrac{1}{2018}$