Đặt \(A=\frac{4}{9.11}+\frac{4}{11.13}+\frac{4}{13.15}+...+\frac{4}{97.99}\)
\(\Rightarrow A=2.\left(\frac{4}{9}-\frac{4}{11}+\frac{4}{11}-\frac{4}{13}+...+\frac{4}{97}-\frac{4}{99}\right)\)
\(\Rightarrow A=2.\left(\frac{4}{9}-\frac{4}{99}\right)=2.\left(\frac{40}{99}\right)=\frac{80}{99}\)
Vậy \(A=\frac{80}{99}\)
\(\frac{4}{9.11}+\frac{4}{11.13}+\frac{4}{13.15}+...+\frac{4}{97.99}=2.\left(\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{97.99}\right)\)
\(=2.\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(\frac{1}{9}-\frac{1}{99}\right)=2.\left(\frac{11}{99}-\frac{1}{99}\right)=2.\frac{10}{99}=\frac{20}{99}\)
