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!

E= 7/4x7 + 7/7x10 =7/10x13+...+ 7/301x304

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

\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{304}\right)\)

\(=\frac{7}{3}\cdot\frac{75}{304}\)

\(=\frac{175}{304}\)

E = \(\frac{7}{4.7}+\frac{7}{7.10}+\frac{7}{10.13}+...+\frac{7}{301.304}\) 

   =\(\frac{7}{3}.\frac{7-4}{4.7}+\frac{7}{3}.\frac{10-7}{7.10}+\frac{7}{3}.\frac{13-10}{10.13}+...+\frac{7}{3}.\frac{304-301}{301.304}\)

 = \(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)=\(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)=\frac{7}{3}.\frac{75}{304}=\frac{175}{304}\)

3/1x4+3/4x7+...+3/97x100

=1-1/4+1/4-1/7+1/7-...+1/97-1/100

=1-1/100

=99/100

đây kô phải toán lớp 1

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

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

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

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

=3/1-3/4+3/4-3/7+3/7-3/10+3/10-3/13

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

=1-1/13

=12/13

3/1x4+3/4x7+3/7x10+3/10x13

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

=1-1/13

=12/13

\(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}\)

\(=3\times\left(\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}\right)\)

\(=3\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)

\(=3\times\left(1-\frac{1}{13}\right)\)

\(=3\times\frac{12}{13}\)

\(=\frac{36}{13}\)

mk đang cần gấp

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

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

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

tại sao lại như thế?

1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100

= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)

= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)

= 1/3(1 - 1/100)

= 1/3*99/100

= 33/100

trả lời 

=33/100

chúc bn

học tốt

Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)

=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)

=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)

=\(\frac{1}{2}-\frac{1}{509}\)

=\(\frac{507}{1018}\)

MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))

làm 2 câu còn lại đi câu đó làm rồi

\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)

Em cảm ơn chị