Đặt biểu thức là A
=> \(A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}.\frac{100}{101}\)
=> \(A=\frac{250}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{3.7}+\frac{5}{7.9}+....+\frac{5}{99.101}\)
\(=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+.....+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+..+\frac{5}{99.101}\)
=\(\frac{5}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+..+\frac{101-99}{99.101}\right)\)
=\(\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{5}{2}\left(1-\frac{1}{101}\right)=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)