Giải:
k) \(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(=\dfrac{4}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}.\dfrac{33}{68}\)
\(=\dfrac{11}{17}\)
l) \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{99.101}{100^2}\)
\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(=\dfrac{\left(1.2.3.....99\right)}{\left(2.3.4.....100\right)}.\dfrac{\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Chúc bạn học tốt!
k) \(\dfrac{4}{2.5} + \dfrac{4}{5.8} + \dfrac{4}{8.11} + ... + \dfrac{4}{65.68}\)
\(= \dfrac{4}{3}\bigg(\dfrac{3}{2.5} + \dfrac{3}{5.8} + ... + \dfrac{3}{65.65}\bigg)\)
\(= \dfrac{4}{3}\bigg(\dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + ... + \dfrac{1}{65} - \dfrac{1}{68}\bigg)\)
\(= \dfrac{4}{3}\bigg(\dfrac{1}{2} - \dfrac{1}{68}\bigg)\)
\(=\dfrac{4}{3}.\dfrac{33}{68}\)
\(= \dfrac{11}{17}\)
l) \(\dfrac{3}{4} . \dfrac{8}{9} . \dfrac{15}{16} . ..... . \dfrac{9999}{10000}\)
\(= \dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)
\(= \dfrac{1.2.3....99}{2.3.4....100} . \dfrac{3.4.5.....101}{2.3.4....100}\)
\(= \dfrac{1}{100} . \dfrac{101}{2}\)
\(= \dfrac{101}{200}\)
