ta nhân 3 cả hai vế, được : 

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)

hay 

\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)

\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)

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.`

Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)

           1/4-1/7 = 3/28 = 3.(1/4.7)

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

A = 3.(1-1/100)

A = 3.(99/100)

A = 297/100

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

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

\(A=\frac{1}{3}.\frac{99}{100}\)

\(A=\frac{33}{100}\)

1) 

A= \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{40}\)

=> A= 27/120

A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

= \(\frac{1}{3}-\frac{1}{40}\)

= \(\frac{37}{120}\)

B = \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{37.40}\)

= \(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

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

= \(\frac{1}{3}.\frac{9}{40}=\frac{3}{40}\)

C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)

= \(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

= \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)

= \(\frac{2}{3}.\frac{9}{40}=\frac{3}{20}\)

1) A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

A = \(\frac{1}{3}-\frac{1}{40}\)

A = \(\frac{37}{120}\)

2) B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{37}-\frac{1}{40}\right)\)

B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

B = \(\frac{1}{3}.\frac{9}{40}\)

B = \(\frac{3}{40}\)

3) C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{37}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\frac{9}{40}\)

C = \(\frac{3}{20}\)

\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)

\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)

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

\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)

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

ai tk mk thì mk tk lại