\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{296.299}\)

=\(\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.7}+\frac{3}{7.11}+...+\frac{3}{296.299}\right)\)

=\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{296}-\frac{1}{299}\right)\)

=\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{299}\right)\)

=\(\frac{1}{3}.\frac{297}{598}\)

=\(\frac{99}{598}\)

3/2.5 + ...+ 3/17 .20

= 3/2 .(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 120)

= 3/2 . (1/2 - 1/20)

= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)

P= 3/ 2.5 + 3/5.8 + 3/8.11 + .... + 3/17.20

P= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + .... + 1 /17 - 1/20

P= 1 / 2 - 1 / 20

P = 9/20

Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)

\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{80.83}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{80}-\frac{1}{83}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{83}\right)\)

\(=\frac{54}{83}\)

\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+............+\frac{4}{80\cdot83}\)

\(=4\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+............+\frac{4}{80\cdot83}\right)\)

\(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...........+\frac{1}{80}-\frac{1}{83}\right)\)

\(=\frac{4}{3}\cdot\frac{81}{166}\)

\(=\frac{54}{83}\)

CHúc các bạn học tôt !!!!!!!!!!!!!!!!!!!

=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*48/98

=1/3*24/49

=8/49

Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

#)Giải :

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)

\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{99}{202}\)

\(\Leftrightarrow A=\frac{33}{202}\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)

A = 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14

A = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14

A = 1/2 - 1/14

A = 7/14 - 1/14

A = 3/7

=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)

           = 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125

Ta có: 3/ 2.5 = 1/2 - 1/5 

          3/5.8  = 1/5 -1/8

          3/ 8.11 = 1/8 -1/11

          ..........................

         3/122 . 125 = 3/122 - 3/125

=> 3B=  1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125

         =  1/2 - 1/125 = 125/250 - 2/250= 123/250

=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250

=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)

           = 2/9.11 + 2/11 .13 +.....+ 2/ 97.99

Ta có: 2/9.11 = 1/9 - 1/11

          2/11.13 = 2/11 -2/ 13

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

         2/97.99 = 1/97 - 1/99

=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99

           = 1/9 -1/99 = 11/99 - 1/99 =10/99

=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99

Vậy B = 5/99

`D=1/2.5+1/5.8+1/8.11+...+1/1979.1982`

`=3/3(1/2.5+1/5.8+1/8.11+...+1/1979.1982)`

`=1/3(3/2.5+3/5.8+3/8.11+...+3/1979.1982)`

`=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/1979-1/1982)`

`=1/3(1/2-1/1982)`

`=1/3(991/1982-1/1982)`

`=1/3 . 495/991`

`=165/991`