b: \(B=\dfrac{5}{1\cdot9}+\dfrac{5}{9\cdot17}+...+\dfrac{5}{793\cdot801}\)
\(=\dfrac{5}{8}\left(\dfrac{8}{1\cdot9}+\dfrac{8}{9\cdot17}+...+\dfrac{8}{793\cdot801}\right)\)
\(=\dfrac{5}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{793}-\dfrac{1}{801}\right)\)
\(=\dfrac{5}{8}\left(1-\dfrac{1}{801}\right)=\dfrac{5}{8}\cdot\dfrac{800}{801}=\dfrac{500}{801}\)
c: \(M=\dfrac{4}{2\cdot4}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{38\cdot40}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{38\cdot40}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{38}-\dfrac{1}{40}\right)\)
\(=2\left[\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{40}\right)-2\cdot\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{40}\right)\right]\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{40}\right)-4\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{40}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
\(=2\left(\dfrac{1}{22}+\dfrac{1}{24}+...+\dfrac{1}{40}\right)\)
=2N
=>\(\dfrac{M}{N}=2\)
