a) \(A=\dfrac{3}{3\cdot9}+\dfrac{3}{9\cdot15}+...+\dfrac{3}{597\cdot603}\)
\(=\dfrac{3}{6}\cdot\left(\dfrac{6}{3\cdot9}+\dfrac{6}{9\cdot15}+...+\dfrac{6}{597\cdot603}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{597}-\dfrac{1}{603}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{603}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{200}{603}\)
\(=\dfrac{100}{603}\)
b) \(B=\dfrac{5}{1\cdot9}+\dfrac{5}{9\cdot17}+...+\dfrac{5}{793\cdot801}\)
\(=\dfrac{5}{8}\cdot\left(\dfrac{8}{1\cdot9}+\dfrac{8}{9\cdot17}+...+\dfrac{5}{793\cdot801}\right)\)
\(=\dfrac{5}{8}\cdot\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{793}-\dfrac{1}{801}\right)\)
\(=\dfrac{5}{8}\cdot\left(1-\dfrac{1}{801}\right)\)
\(=\dfrac{5}{8}\cdot\dfrac{800}{801}\)
\(=\dfrac{500}{801}\)
