Sửa đề: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

Ta có: \(\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

\(=\left(1+\frac{2019}{2}\right)+\left(1+\frac{2018}{3}\right)+\cdots+\left(1+\frac{1}{2020}\right)+1\)

\(=\frac{2021}{2}+\frac{2021}{3}+\cdots+\frac{2021}{2021}=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

Ta có: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

=>x=2021

C

C

Dễ thấy A chia hết cho 10 nên A có tận cùng là 0

còn 1x 3 x 5 x... x 2021 là một số lẻ và chia hết cho 5 nên có tận cùng là 5

B/A

\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)

\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)

Sửa đề; \(\frac{x+1}{2021}+\frac{x+2}{2020}+\frac{x+3}{2019}=-3\)

Ta có: \(\frac{x+1}{2021}+\frac{x+2}{2020}+\frac{x+3}{2019}=-3\)

=>\(\left(\frac{x+1}{2021}+1\right)+\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+3}{2019}+1\right)=-3+3=0\)

=>\(\frac{x+2022}{2021}+\frac{x+2022}{2020}+\frac{x+2022}{2019}=0\)

=>x+2022=0

=>x=-2022

khó hiểu vcl

đúng lun ko hiểu một chút nào