Đặt A = \(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=10\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2015.2020}\right)\)
\(=\frac{10}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{2020}\right)\)\(=\frac{2}{5}-\frac{2}{2020}\)
\(=\frac{2}{5}-\frac{1}{1010}\)\(=\frac{404}{1010}-\frac{1}{1010}\)\(=\frac{403}{1010}\)
Vậy giá trị của biểu thức đã cho là 403/1010
\(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=2.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(\frac{2}{5}-\frac{1}{1010}\)
Tính nốt nha
