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11 tháng 4 2016

Gọi biểu thức đó là A

Ta có: \(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{17.20}\)

\(A:4.3=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(A:4.3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(A:4.3=\frac{1}{2}-\frac{1}{20}\)

\(A:4.3=\frac{9}{20}\)

\(A=\frac{3}{5}\)

11 tháng 4 2016

k mình nha

28 tháng 3 2017

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

28 tháng 3 2017

=\(\frac{3}{20}=0,15\)

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

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

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

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

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

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

12 tháng 5 2017

A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

12 tháng 5 2017

sai ùi 

11 tháng 3 2017

\(\frac{4}{2\cdot5}\)+\(\frac{4}{5\cdot8}\)+\(\frac{4}{8\cdot11}\)+.......+\(\frac{4}{8\cdot83}\)=\(\frac{4}{3}\) (\(\frac{3}{2\cdot5}\)+\(\frac{3}{5\cdot8}\) +......+\(\frac{3}{80\cdot83}\) )

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

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

=\(\frac{4}{3}\)*\(\frac{81}{166}\) 

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

24 tháng 3 2016

\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)

\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)

\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)

\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)

=> \(x=2\)

24 tháng 3 2016

\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)

\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)

\(=>x=1\)

9 tháng 8 2015

S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48

S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )

S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )

S = 4/3 . ( 1/2 - 1/68 )

S = 4/3 . 33/68

S = 11/17

12 tháng 4 2018

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

25 tháng 4 2018

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

\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.........+\frac{1}{65.68}\right)\)

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

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

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

\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{33}{68}\right)\)

\(A=\frac{11}{17}\)

Vậy A = \(\frac{11}{17}\)

Chúc bạn học tốt!

16 tháng 5 2016

S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)

   = 1/3 . (1/2 - 1/20)

   = 1/3 . 9/20

   = 3/20

16 tháng 5 2016

\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)

\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(S=\frac{9}{20}:3=\frac{3}{20}\)

10 tháng 5 2019

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

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

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

\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)

\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)

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

\(A=\frac{11}{17}\)

~ Hok tốt ~

10 tháng 5 2019

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

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

       \(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)