b) Ta có: \(B=\dfrac{2020}{1}+\dfrac{2019}{2}+\dfrac{2018}{3}+...+\dfrac{1}{2020}\)
\(=\left(1+\dfrac{2019}{2}\right)+\left(1+\dfrac{2018}{3}\right)+...+\left(1+\dfrac{1}{2020}\right)+1\)
\(=\dfrac{2021}{2}+\dfrac{2021}{3}+...+\dfrac{2021}{2020}+\dfrac{2021}{2021}\)
\(=2021\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2020}+\dfrac{1}{2021}\right)\)
Ta có: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{\dfrac{2020}{1}+\dfrac{2019}{2}+\dfrac{2018}{3}+...+\dfrac{1}{2020}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2020}+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
