\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)

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

\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)

\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)

\(A=\frac{3}{1}-\frac{3}{100}\)

\(A=\frac{297}{100}\)

\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)

\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)

\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)

\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(=3\cdot\left(1-\dfrac{1}{16}\right)\)

\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)

\(=3\cdot\dfrac{15}{16}\)

\(=\dfrac{45}{16}\)

ngu :3

3/1.4+3/4.7+3/7.10+3/10.13

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

=1 - 1/13

=12/13

#)Trả lời :

Gọi tổng trên là A 

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}\)

\(A=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}\)

\(A=3-\frac{3}{13}\)

\(A=2\frac{10}{13}=\frac{36}{13}\)

               #~Will~be~Pens~#

  3/1.4 + 3/4.7 + .. +3/13.16

= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16

= 1/1 - 1/16

= 15/16

\(=\frac{15}{16}\)

đúng cho mk nha Minh Thư Nguyễn

a 

9/38 

chuc ban hoc tot

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)

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

Ta thấy :

 \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

\(.........\)

\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)

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

\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)

đáp án = \(\frac{297}{100}\)

đúng không?

kết bạn với mh nha

\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)

\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)

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

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

\(=3.\frac{99}{100}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2008.2011}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2008}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}\)

\(=\frac{2011}{2011}-\frac{1}{2011}\)

\(=\frac{2010}{2011}\)

Chúc bạn học tốt !!!! 

Đặt: A= \(\frac{3}{1\times4}\)+ \(\frac{3}{4\times7}\)+ \(\frac{3}{7\times10}\)+...+ \(\frac{3}{2005\times2008}\)+ \(\frac{3}{2008\times2011}\).

A= \(\frac{3}{1}\)- \(\frac{3}{4}\)+ \(\frac{3}{4}\)- \(\frac{3}{7}\)+ \(\frac{3}{7}\)- \(\frac{3}{10}\)+...+ \(\frac{3}{2005}\)- \(\frac{3}{2008}\)+ \(\frac{3}{2008}\)- \(\frac{3}{2011}\).

A= 3- \(\frac{3}{2011}\).

A= \(\frac{6033}{2011}\)- \(\frac{3}{2011}\).

A= \(\frac{6030}{2011}\).

Vậy A= \(\frac{6030}{2011}\).

\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2024}=\dfrac{1}{4}-\dfrac{1}{2024}=\dfrac{505}{2024}\)

\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{1}{2021.2024}\)

=\(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{2021}-\)\(\dfrac{1}{2024}\)

=\(\dfrac{1}{4}-\dfrac{1}{2024}\)

=\(\dfrac{505}{2024}\)