\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=3.\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

\(A=\dfrac{3^2}{1\times4}+\dfrac{3^2}{4\times7}+\dfrac{3^2}{7\times10}+\dfrac{3^2}{10\times13}+\dfrac{3^2}{13\times16}...+\dfrac{3^2}{97\times100}\)

\(=3\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16} +...+\dfrac{3}{97\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

\(=2\dfrac{97}{100}\)

Vậy \(A=2\dfrac{97}{100}\)

\(l=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{97.100}\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}-\dfrac{1}{300}< \dfrac{1}{3}\left(đpcm\right)\)

\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy \(S=\dfrac{99}{100}\)

Chúc bạn học tốt!!!

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{97}-\dfrac{1}{100}\)

\(S=1-\dfrac{1}{100}\)

\(S=\dfrac{99}{100}\)

\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)

=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)

=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

=> \(2.\left(1-\dfrac{1}{100}\right)\)

=>\(2\).\(\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)

= \(2.\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)

\(=\dfrac{5^3\cdot2^3+2\cdot5^3+5^3}{55}=\dfrac{5^3\left(2^3+2+1\right)}{55}=5^2=25\)

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`

b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)

\(=\dfrac{13}{276}\)

\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)

\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)

\(2.5+1.4-1\dfrac{3}{7}=\dfrac{5}{2}+\dfrac{7}{5}-\dfrac{10}{7}=\dfrac{5\cdot35}{2\cdot35}+\dfrac{7\cdot14}{5\cdot14}-\dfrac{10\cdot10}{7\cdot10}=\dfrac{175}{70}+\dfrac{98}{70}-\dfrac{100}{70}=\dfrac{173}{70}\)

\(=3,9+\dfrac{10}{7}=\dfrac{39}{10}+\dfrac{10}{7}=\dfrac{173}{70}\)

2,5 + 1,4 - \(1\dfrac{3}{7}\)

=3,9 + \(\dfrac{10}{7}\)

=\(\dfrac{373}{70}\)