A=4.7+7.10+10.13+...+205.208

A.9=4.7.9+7.10.9+10.13.9+...+205.208.9

A.9=4.7.(10-1)+7.10.(13-4)+...+205.208.(211-202)

A.9=4.7.10-1.4.7+7.10.13-4.7.10+...+205.208.211-202.205.208

A.9=-1.4.7+205.208.211

A.9=8997012

A=8997012:9=999668

Ta có:

Đặt \(A=\)\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{604.607}< \dfrac{1}{2}\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{604}-\dfrac{1}{607}\right)< \dfrac{1}{2}\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{607}\right)< \dfrac{1}{2}\)

Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) nên \(\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{607}\right)< \dfrac{1}{2}\) 

Vậy \(A< \dfrac{1}{2}\)

............................... =) A < 1/2

\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)

`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`

`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`

`3A = 3/14`

`A = 1/14.`

S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64

1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2

4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7

7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10

10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13

.......................................................................

61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64

Cộng vế với vế ta có:

1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67

9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576

  1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9

1.4 + 4.7 + 7.10 + ... + 61.64 = 29064 

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

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

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

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

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

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

\(=\dfrac{3}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{2021.2024}\right)\)

\(=1.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\)

\(=1.\left(\dfrac{1}{4}-\dfrac{1}{2024}\right)\)

\(=1.\left(\dfrac{506-1}{2024}\right)\)

\(=1.\dfrac{505}{2024}\)

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

a) \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

b) \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(B=\frac{1}{4}-\frac{1}{76}\)

\(B=\frac{19}{76}-\frac{1}{76}=\frac{18}{76}=\frac{9}{38}\)

a, \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{4}{13}\)

b, \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(B=\frac{1}{4}-\frac{1}{76}\)

\(B=\frac{9}{38}\)

Ta có: \(c=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{37\cdot40}\)

\(\Leftrightarrow3c=3\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{37\cdot40}\right)\)

\(\Leftrightarrow3c=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{37\cdot40}\)

Mà \(\frac{3}{4\cdot7}=\frac{1}{4}-\frac{1}{7}\)

\(\frac{3}{7\cdot10}=\frac{1}{7}-\frac{1}{10}\)

...

\(\Leftrightarrow3c=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{37\cdot40}\)

\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{40}\)

Ta thấy ngoại trừ hai phân số đầu tiên và cuối cùng thì tất cả các phân số còn lại đều có 1 phân số có cùng giá trị tuyệt đối nhưng ngược dấu đứng cạnh, mà tổng hai số ngược dấu bằng 0 nên ta nhóm các phân số ngược dấu thì được:

\(3c=\frac{1}{4}-\frac{1}{40}\Leftrightarrow c=\left(\frac{1}{4}-\frac{1}{40}\right)\cdot\frac{1}{3}\)

\(=\frac{9}{40}\cdot\frac{1}{3}=\frac{3}{40}=\frac{9}{120}< \frac{40}{120}\)

Mà \(\frac{40}{120}=\frac{1}{3}\Rightarrow c< \frac{1}{3}\)

\(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}\)

\(=2\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}\right)\)

\(=2\left[\frac{1}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right)\right]\)

\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\right]\)

\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\right]\)

\(=2\left[\frac{1}{3}\cdot\frac{3}{16}\right]\)

\(=2\cdot\frac{1}{16}\)

\(=\frac{2}{16}=\frac{1}{8}\)

Ta có : 

\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\) 

\(=\)\(2\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\right)\)

\(=\)\(\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)

\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(=\)\(\frac{2}{3}.\frac{3}{16}\)

\(=\)\(\frac{1}{8}\)

Chúc bạn học tốt ~ 

\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)

= \(\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)

= \(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

= \(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\)

= \(\frac{2}{3}.\frac{3}{16}\)

= \(\frac{1}{8}\)

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~#

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\)

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

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

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