Đặt A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}=\frac{8}{25}\)
Vậy \(A>\frac{8}{25}\left(\text{ĐPCM}\right)\)