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\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)=\dfrac{4}{3}\cdot\dfrac{303}{610}=\dfrac{202}{305}\)

24 tháng 2 2022

\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{302.305}\)

\(=4\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}.\dfrac{303}{610}\\ =\dfrac{202}{305}\)

19 tháng 3 2018

\(=\frac{3}{4}\cdot\left(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{62.65}\right)\)

\(=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\)

\(=\frac{1}{2}-\frac{1}{65}\)

\(=\frac{63}{130}\)

19 tháng 3 2018

Đặt A=4/2.5+4/5.8+4/8.11+...+4/62.65.Ta có A=4.(1/2.5+1/5.8+1/8.11+...1/62.65)=4/3.(3/2.5+3/5.8+3/8.11+...+3/62.65)                                         =4/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+3/62-3/65)=4/3.(1/2-1/65)=4/3.63/130=42/56                                                                                            Vậy A=42/56

25 tháng 3 2017

 bằng 3.141939849 kết bạn với mình nhé

19 tháng 4 2021

Ta có : B = 4 / 2. 5 + 4 / 5 . 8 + 4 / 8 . 11 + 4 / 11 . 14

               = 4 ( 1 / 2 . 5 + 1 / 5 . 8 + 1 / 8 . 11 + 1 / 11 . 14 )

              = 4 . [ 1/3 ( 1 /2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/ 11 + 1/11 - 1/14 )]

              = 4 / 3 ( 1/ 2 - 1/14 )

             = 4/3 . 3/7

             = 4 / 7 

Vậy B = 4 / 7 

14 tháng 5 2016

M = 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20

M= 4/3 . (1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)

M= 4/3 . (1/2 - 1/20)

M= 4/3 . (10/20 - 1/20)

M= 4/3 . 9/20

M= 3/5

k nha

Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)

\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)

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

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)

24 tháng 7 2021

Thank bạn!

25 tháng 8 2023

Sửa đề:

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

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

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

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

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

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

25 tháng 8 2023

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 = \(\dfrac{11}{17}\)

8 tháng 1 2018

\(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+\dfrac{4}{8\cdot11}+...+\dfrac{4}{65\cdot68}\\ =\dfrac{4}{3}\cdot\dfrac{3}{2\cdot5}+\dfrac{4}{3}\cdot\dfrac{3}{5\cdot8}+\dfrac{4}{3}\cdot\dfrac{3}{8\cdot11}+...+\dfrac{4}{3}\cdot\dfrac{3}{65\cdot68}\\ =\dfrac{4}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{65\cdot68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\dfrac{33}{68}\\ =\dfrac{11}{17}\)

26 tháng 4 2017

3/4 . M = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20

3/4 . M = 1/2 - 1/20

3/4 .M = 9/10

M = 9/10 : 3/4

M = 3/5

 Vậy M = 3/5

Nhớ tk m nha

28 tháng 4 2018

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

7A = \(\dfrac{4.3}{2.5}\) + \(\dfrac{4.3}{5.8}\) + \(\dfrac{4.3}{8.11}\) + ... + \(\dfrac{4.3}{65.68}\)

7A = 4 (\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{65.68}\))

7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + ... + \(\dfrac{1}{65}\) - \(\dfrac{1}{68}\))

7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

7A = 4 . \(\dfrac{33}{68}\) = \(\dfrac{33}{17}\)

A = \(\dfrac{33}{17}\) : 7

=> A = \(\dfrac{33}{119}\)

28 tháng 4 2018

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

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

\(=\dfrac{4}{3}\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+...+\dfrac{68-65}{65.68}\right)\)

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

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)=\dfrac{4}{3}.\dfrac{33}{68}=\dfrac{11}{17}\)

7 tháng 8 2016

\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\frac{99}{202}\)

\(=\frac{66}{101}\)

7 tháng 8 2016

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\) 

\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)

\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\) 

\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\) 

\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)