\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
(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!!!
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{97}-\dfrac{1}{100}\)
\(S=1-\dfrac{1}{100}\)
\(S=\dfrac{99}{100}\)
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=\(\dfrac{99}{100}\)
Chúc bạn học tốt 
Giải:
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}.\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}.\)
\(S=1+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+...+\left(\dfrac{1}{97}-\dfrac{1}{97}\right)-\dfrac{1}{100}.\)
\(S=1+0+0+...+0-\dfrac{1}{100}.\)
\(S=1-\dfrac{1}{100}=\dfrac{99}{100}.\)
Vậy \(S=\dfrac{99}{100}.\)
