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4/2.4+4/4.6+4/6.8+.....+4/1008.2010
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29 tháng 4 2015
A=4/2.4+4/4.6+4/6.8+...+4/2008.2010
=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
Vậy A=1004/1005
29 tháng 4 2015
100% giải đúng đầu tiên:
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
12 tháng 2 2018
=> K : 2 = \(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}=\frac{502}{1005}\)
\(\Rightarrow K=\frac{1004}{1005}\)
Vậy \(K=\frac{1004}{1005}\)
1 tháng 5 2016
F=2 .(1/2-1/4+1/4-1/6+......+1/2008 - 1/2010)
= 2.(1/2-1/2010)
= 2. 502/1005
= 1004/1005
L3
4
5 tháng 7 2021
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
SI
5 tháng 7 2021
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
25 tháng 4 2018
S = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2016.2018}\)
S = \(2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{2018}\right)\)
S = \(2.\frac{504}{1009}\)= \(\frac{1008}{1009}\)
Vậy S = \(\frac{1008}{1009}\).
~~~
Nếu có sai sót gì thì giúp đỡ tớ nha :3
#Sunrise
NQ
0
3 tháng 5 2015
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
TN
2
29 tháng 4 2016
K = 2( 2/2.4 + 2/4.6 +......+ 2/2008.2010)
K = 2( 1/2 - 1/4 + 1/4 - 1/6 +......+ 1/2008 - 1/2010)
K = 2( 1/2 - 1/2010)
K = 2 . 1004/2010
K = 1004/1005
Ai k mk mk k lại
29 tháng 4 2016
K=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
K=\(\frac{4}{2}.\frac{2}{2.4}+\frac{4}{2}.\frac{2}{4.6}+...+\frac{4}{2}.\frac{2}{2008.2010}\)
K=\(\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
K=\(\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{2008}-\frac{1}{2010}\right)\)
K=\(\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
K=\(\frac{4}{2}.\frac{502}{1005}\)
K=\(\frac{1004}{1005}\)
TN
1
TN
29 tháng 4 2016
\(K=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2\times\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)
1008 các bạn sửa lại giúp mink là 2008
\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+............+\dfrac{4}{2008.2010}\)
\(=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+......+\dfrac{2}{2008.2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)