theo bài ra ta có:

\(E=\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{74.77}\\ \Rightarrow\dfrac{1}{5}E=\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{74.77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{74}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{7}{77}-\dfrac{1}{77}=\dfrac{6}{77}\\ \Rightarrow E=\dfrac{6}{77}.5\\ E=\dfrac{30}{77}\)

5 .\((\)\(\dfrac{3}{11.14}+\dfrac{3}{14.17}+...+\dfrac{3}{74.77}\))

= 5. (\(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{74}-\dfrac{1}{77}\))

= 5.(\(\dfrac{1}{11}-\dfrac{1}{77}\))

= 5. \(\dfrac{6}{77}\)

= \(\dfrac{30}{77}\)

E = \(\dfrac{30}{77}\) thì phải

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}\)

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(A=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{21}\)

\(A=\frac{1}{3}-\frac{1}{21}\)

\(A=\frac{2}{7}\)

\(B=\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{68.71}\)

\(=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+...+\frac{1}{68}-\frac{1}{71}\)

\(=\frac{1}{11}-\frac{1}{71}\)

\(=\frac{71}{781}-\frac{11}{781}\)

\(=\frac{60}{781}\)

\(\frac{3}{15}\cdot G=\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{68\cdot71}\)

\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)

\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{71}\)

\(G=\frac{60}{781}\cdot\frac{15}{3}\)

\(G=\frac{300}{781}\)

ta có :\(\frac{3}{15}G=\left(\frac{15}{11.14}+\frac{15}{14.17}+...+\frac{15}{68.71}\right)\)

\(\frac{3}{15}G=\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{68.71}\)

\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)

\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{71}=\frac{71}{781}-\frac{11}{781}=\frac{60}{781}\)

\(=>G=\frac{60}{781}:\frac{3}{15}=\frac{900}{2343}\)

vậy G =900/2343

\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+...+\frac{15}{72.75}\)

\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{72.75}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+...+\frac{1}{72}-\frac{1}{75}\right)\)\(=5\left(\frac{1}{11}-\frac{1}{75}\right)\)

\(=\frac{64}{165}\)

có ai bt bấm tổng xích ma bài này ko, chỉ mk vs !

\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{68.71}\)

\(=5\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{68}-\dfrac{1}{71}\right)\)

\(=5\left(\dfrac{1}{11}-\dfrac{1}{71}\right)\)

\(=5.\dfrac{60}{781}\)

\(=\dfrac{300}{781}\)

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

=\(\frac{3}{20}=0,15\)

\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

sai ùi 

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}.\frac{9}{20}\)

\(A=\frac{3}{20}\)

Vì \(\frac{3}{20}< 1\Rightarrow A< 1\)

\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)

\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)

\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)

\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)

=> \(x=2\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)

\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)

\(=>x=1\)

Ta có : \(\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-......-\frac{15}{47.45}\)

\(=\frac{3}{8}-\left(\frac{15}{8.11}+\frac{15}{11.14}+\frac{15}{14.17}+......+\frac{15}{47.50}\right)\)

\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{11}{47}-\frac{1}{50}\right)\)

\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{50}\right)\)

\(=\frac{3}{8}-\frac{1}{8}+\frac{1}{50}\)

\(=\frac{1}{4}+\frac{1}{50}=\frac{27}{100}\)