Ta có:

A = \(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{301.304}\)

A = 5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))

3A = 3.5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))

3A = 5. (\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\))

3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))

3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))

3A = 5. ( \(\frac{1}{4}-\frac{1}{304}\))

3A = \(\frac{5.75}{304}\)

3A = \(\frac{375}{304}\)

A= \(\frac{125}{304}\) . Vậy: A = \(\frac{125}{304}\)

Chúc bạn học tốt! Tick cho mình nhé!

\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)

\(=\frac{5}{3}.\frac{75}{304}\)

\(=\frac{125}{304}\)

\(\frac{5}{4×7}+\frac{5}{7×10}+\frac{5}{10×13}+...+\frac{5}{301×304}\)

\(=\frac{5}{4}-\frac{5}{7}+\frac{5}{7}-\frac{5}{10}+\frac{5}{10}-\frac{5}{13}+...+\frac{5}{301}-\frac{5}{304}\)

\(=\frac{5}{4}-\frac{5}{304}\)

\(=\frac{380}{304}-\frac{5}{304}\)

\(=\frac{375}{304}\)

Cbht

\(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{301.304}\)

\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)

\(=\frac{5}{3}.\frac{75}{304}\)

\(=\frac{125}{304}\)

\(C=\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{2020+2023}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{2020.2023}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{3}.\dfrac{2019}{8092}\)

\(=\dfrac{673}{8092}\)

=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16

=1-1/16=15/16

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)

\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)

\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)

\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)

\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)

1/1.4+1/4.7+1/7.10+1/10.13+1/13.16

=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)

=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)

=1/3.(1/1-1/16)

=1/3.(16/16-1/16)=1/3.15/16=5/16