EC
10
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(\Rightarrow A=\frac{33}{50}\)
AT
2
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
ML
12 tháng 8 2015
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.....+\frac{2}{601.604}=\frac{2}{3}\cdot\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{601.604}\right)=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{601}-\frac{1}{604}\right)\)=\(\frac{2}{3}\cdot\left(1-\frac{1}{604}\right)=\frac{2}{3}\cdot\frac{603}{604}=\frac{201}{302}\)
T
9 tháng 9 2018
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
ND
2
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 \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)
\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)
\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)
AT
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 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)
NT
0
TT
1
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 !!!
\(B=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-...-\frac{2}{100}\right)\)
\(B=\frac{1}{3}.\left(2-\frac{2}{100}\right)=\frac{1}{3}.\frac{99}{50}==\frac{33}{50}\)