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tui chỉ làm phần b thôi há !

B=\(\frac{3}{4}\)+\(\frac{27}{28}\)+\(\frac{69}{70}\)+...+\(\frac{867}{868}\)=\(\frac{4-1}{4}\)+\(\frac{28-1}{28}\)+\(\frac{70-1}{70}\)+...+\(\frac{868-1}{868}\)

= 1+1+..+1 -(\(\frac{1}{4}\)+\(\frac{1}{28}\)+\(\frac{1}{70}\)+...+\(\frac{1}{868}\)) = 10 - \(\frac{1}{3}\)(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+...+\(\frac{3}{28.31}\))

=10-\(\frac{1}{3}\)(1-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{10}\)+...+\(\frac{1}{28}\)-\(\frac{1}{30}\))=10-\(\frac{1}{3}\)(1-\(\frac{1}{30}\))=10-\(\frac{1}{3}\).\(\frac{29}{30}\)=10-\(\frac{29}{30}\)=\(\frac{271}{30}\)

=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{28}-\frac{2}{31}\right)\)

=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{31}\right)=\frac{20}{31}\)

Bấm đúng cho tui, đi mà. CHÚC BẠN HỌC GIỎI

bài giải đó là sai giả như vầy nè

\(=\frac{1}{3}\cdot2\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{31}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{31}\right)\)

=\(\frac{2}{3}\cdot\frac{30}{31}=\frac{20}{31}\)

Trả lời

3/1.4+3/4.7+3/7.10

=3/1.3/10

=3/10

Chúc bạn học tốt #

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

tớ ko biết

S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)

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

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

Trả lời

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{4.7}+...+\frac{3}{97.100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

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

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

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

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

    \(\frac{297}{100}\)

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