1/1.2 + 1/2.4 + 1/4.6 + ...+1/2012.2014
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
N
4
HT
4 tháng 8 2015
A = \(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2012.2014}\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2012}-\frac{1}{2014}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014}\right)\)
A = \(\frac{1}{2}.\frac{503}{1007}\)
A = \(\frac{503}{2014}\)
4 tháng 8 2015
2A=\(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2014-2012}{2012.2014}\)
\(2A=\frac{4}{2.4}-\frac{2}{2.4}+\frac{6}{4.6}-\frac{4}{4.6}+...+\frac{2014}{2012.2014}-\frac{2012}{2012.2014}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)
\(2A=\frac{1}{2}-\frac{1}{2014}=\frac{503}{1007}\Rightarrow A=\frac{503}{2014}\)
NQ
1
VD
21 tháng 4 2016
a,A=4/2.4+4/4.6+4/6.8+......+4/2012.2014
\(\Rightarrow\frac{1}{2}A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{2012\cdot2014}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2014}\)
\(\Rightarrow A=1-\frac{1}{1007}\)
\(\Rightarrow A=\frac{1006}{1007}\)
T
1
IS
24 tháng 3 2017
a, 1/1.2+1/2.3+1/3.4+...+1/999.1000
= 1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000
= 1/1-1/1000
= 999/1000
b, 1/2.4+1/4.6+1/6.8+1/8.10
= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10
= 1/2-1/10
= 4/10 =2/5
20 tháng 4 2019
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}\)
\(=\frac{2007}{1008}\)
KT
20 tháng 4 2019
giải:
4/2.4+4/4.6+4/6.8+...+4/2012.2014+4/2014.2016
=2.(2/2.4+2/4.6+2/6.8+...+2/2012.2014+2/2014.2016
=2.(1/2-1/4+1,4-1/6+1/6-1/8+...+1/2012-1/2014+1/2014-1/2016)
=2.(1/2-1/2016)
=2.1007/2016
=1007/1008
xong rùi đó
LH
8 tháng 3 2020
\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
LH
8 tháng 3 2020
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
S = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}\)\(-\frac{1}{2014}\)
S = \(\frac{1}{1}-\frac{1}{2014}\)
S = \(\frac{2014}{2014}-\frac{1}{2014}\)
S = \(\frac{2013}{2014}\)
\(\frac{1}{1.2}+\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2012.2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)
\(=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{2012}-\frac{1}{2012}\right)-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)