Question

Tính B

B=\(\frac{1}{2019}+\frac{2}{2019}+\frac{3}{2019}+\frac{4}{2019}+...+\frac{2019}{2019}\)

Answer 1

\(B=\frac{1}{2019}+\frac{2}{2019}+\frac{3}{2019}+...+\frac{2019}{2019}\)

\(=\frac{1+2+3+...+2019}{2019}\)

\(=\frac{\left(2019+1\right).\left[\left(2019-1\right)+1\right]:2}{2019}\)

\(=\frac{2039190}{2019}\)

\(=1010\)

Answer 2

#)Giải :

\(B=\frac{1}{2019}+\frac{2}{2019}+\frac{3}{2019}+...+\frac{2019}{2019}\)

\(B=\frac{1+2+3+...+2018+2019}{2019}\)

\(B=\frac{\frac{\left(2019+1\right)\times2019}{2}}{2019}\)

\(B=\frac{2039190}{2019}\)

Answer 3

\(B=\frac{1}{2019}+\frac{2}{2019}+\frac{3}{2019}+\frac{4}{2019}+...+\frac{2019}{2019}\)

\(B=\frac{1+2+3+4+...+2019}{2019}\)

\(B=\frac{\frac{\left(2019+1\right)\times2019}{2}}{2019}\)

\(B=\frac{2039190}{2019}\)

\(B=1010\)

Answer 4

\(B= 2019 1 ​ + 2019 2 ​ + 2019 3 ​ +...+ 2019 2019 ​ B=\frac{1+2+3+...+2018+2019}{2019}B= 2019 1+2+3+...+2018+2019 ​ B=\frac{\frac{\left(2019+1\right)\times2019}{2}}{2019}B= 2019 2 (2019+1)×2019 ​ ​ B=\frac{2039190}{2019}B= 2019 2039190 ​ \)

#YTB

Answer 5

\(B=\frac{1}{2019}+\frac{2}{2019}+...+\frac{2019}{2019}\)

\(=\frac{1+2+...+2019}{2019}\)

\(=\frac{\left(2019+1\right)\times\left[\left(2019-1\right)\div1+1\right]\div2}{2019}\)

\(=\frac{2020\times2019\div2}{2019}\)

\(=\frac{1010.2019}{2019}=1010\)