Question

tính giá trị biểu thức\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2019}{1}+\frac{2019}{2}+\frac{2017}{3}+...+\frac{1}{2019}}\)

Answer 1

Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)

\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)

\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)