tính các tổng sau : \(F=\frac{4}{2.4}\)+ \(\frac{4}{4.6}\)+ \(\frac{4}{6.8}\)+ ... + \(\frac{4}{2008.2010}\)
K
Khách
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LC
7
VQ
27 tháng 3 2015
K = 4/2 - 4/4 + 4/4 - 4/6 + ....... + 4/2008 - 4/2010
K = 4/2 - 4/2010
K = 4016/2010 = 1/1003/1005
ND
3
TM
10 tháng 7 2016
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
Có gì ko hiểu thì cứ hỏi mình nha :)
10 tháng 7 2016
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.2\frac{2}{4}+2.2\frac{2}{4.6}+2.2\frac{2}{6.8}+...+2.2\frac{2}{2008.2010}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}\)
\(=\frac{1004}{1005}\)
TK
3
MU
7 tháng 4 2016
Đặt A= \(\frac{4}{2.4}\)+\(\frac{4}{4.6}\)+\(\frac{4}{6.8}\)+...+\(\frac{4}{2008.2010}\)
A= 2(\(\frac{2}{2.4}\)+\(\frac{2}{4.6}\)+\(\frac{2}{6.8}\)+...+\(\frac{2}{2008.2010}\))
A=2(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\))
A=2(\(\frac{1}{2}-\frac{1}{2010}\))
A=2.\(\frac{502}{1005}\)
A=\(\frac{1004}{1005}\)
7 tháng 4 2016
Mình ko ghi lai đề nha
4/2.4/4+4/4.4/6+......+4/2008.4/2010=4/2.4/2010=4/1005
Mình ko bt đúng ko nữa nha
NA
7
6 tháng 6 2015
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)
6 tháng 6 2015
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)
=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
=\(2\left(1-\frac{1}{1005}\right)\)
=\(2.\frac{1004}{1005}\)
=\(\frac{2008}{1005}\)
P/s: Không biết đúng không nữa, làm đại ^.^
27 tháng 4 2017
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=2.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\)
=2.(\(\frac{1}{2}-\frac{1}{2010}\)) = 2.(\(\frac{1005}{2010}-\frac{1}{2010}\))
=2.\(\frac{502}{1005}\)
=\(\frac{1004}{1005}\)
27 tháng 4 2017
\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1005}{2010}-\frac{1}{2010}\right)\)
\(=2\cdot\frac{1004}{2010}\)
\(=\frac{1004}{1005}\)
\(k\)\(mk\)\(nha\)\(bn\)
LM
21 tháng 8 2016
\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)
LM
21 tháng 8 2016
\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)
\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)
\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)
23 tháng 4 2017
a,\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\) =\(\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\) =\(\left(3-\frac{31}{12}\right):\left(4-\frac{1}{3}\right)\) =\(\frac{5}{12}:\frac{11}{3}\) =\(\frac{5}{44}\) b, F=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.......+\frac{4}{2008.2010}\) =\(2.\left(1-\frac{2}{2010}\right)\) =\(2.\frac{1004}{1005}\) =\(\frac{2008}{1005}\)
NH
5 tháng 5 2019
\(S=\frac{10}{2.4}+\frac{10}{4.6}+...+\frac{10}{2008.2010}\)
\(S=5\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(S=5\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(S=5\left(\frac{1}{2}-\frac{1}{2010}\right)=5.\frac{502}{1005}=\frac{502}{201}\)
5 tháng 5 2019
S=\(\frac{10}{2}-\frac{10}{4}+\frac{10}{4}-\frac{10}{6}+....\frac{10}{2008}-\frac{10}{2010}\)
S=\(\frac{10}{2}-\frac{10}{2010}=5-\frac{1}{201}=\frac{5}{1}-\frac{1}{201}=\frac{1005}{201}-\frac{1}{201}=\frac{1004}{201}\)
14 tháng 8 2016
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A=1-\frac{1}{2010}\)
\(A=\frac{2009}{2010}\)
Ta có:F=4/2.4+4/4.6+4/6.8+...+4/2008.2010
=4/2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
=2.(1/2-1/4+1/4-1/6+1/6-1/8+....+1/2008-1/2010)
=2.(1/2-1/2010)
=2.502/1005
=1004/1005
Mình chắc luôn đó, mình làm bài này rồi!