\(A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+...+\dfrac{1}{95}.\dfrac{1}{99}\)
\(=\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{95.99}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=\dfrac{8}{99}\)
Vậy \(A=\dfrac{8}{99}\)
\(A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{15}+...+\dfrac{1}{95}.\dfrac{1}{99}\)
\(4A=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{95.99}\)
\(4A=\dfrac{7-3}{3.7}+\dfrac{11-7}{7.11}+\dfrac{15-11}{11.15}+...+\dfrac{99-95}{95.99}\)
\(4A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{99}\)
\(4A=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{32}{99}:4=\dfrac{8}{99}\)
Vậy:...
