\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.....+\frac{2}{601.604}\)tính 

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12 tháng 8 2015

C = 2/1.4 + 2/4.7 + 2/7.10 + .... + 2/601.604

C = 2/3 . ( 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/601.604 )

C = 2/3 . ( 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/601 - 1/604 )

C = 2/3 . ( 1 - 1/604 )

C = 2/3 . 603/604

C = 201/302

12 tháng 8 2015

\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.....+\frac{2}{601.604}=\frac{2}{3}\cdot\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{601.604}\right)=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{601}-\frac{1}{604}\right)\)=\(\frac{2}{3}\cdot\left(1-\frac{1}{604}\right)=\frac{2}{3}\cdot\frac{603}{604}=\frac{201}{302}\)

\(B=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{2}{3}\left(1-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

24 tháng 8 2015

\(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-...-\frac{2}{100}\right)\)

\(B=\frac{1}{3}.\left(2-\frac{2}{100}\right)=\frac{1}{3}.\frac{99}{50}==\frac{33}{50}\)

11 tháng 9 2016

\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{33}{50}\)

11 tháng 9 2016

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{2}{3}.\frac{99}{100}\)

\(\Rightarrow A=\frac{33}{50}\)

23 tháng 11 2016

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

23 tháng 11 2016

A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)

A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)

 

6 tháng 10 2019

Sai đề : \(\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

Chúc bạn học tốt !!!

1 tháng 10 2016

anh gửi chơi hay thật vậy

29 tháng 6 2017

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)

29 tháng 6 2017

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}.\frac{18}{19}\)

\(A=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)

\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}.\frac{5}{24}\)

\(B=\frac{5}{48}\)