C = 4/2.4+4/4.6+4/6.8+ ...... + 4/2008.2010
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
K=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
K=2.(4-2/2.4+6-4/4.6+8-6/6.8+...+2010-2008/2008.2010)
K=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
K=2.(1.2-1.2010)
K=2.502/1005
K=1004/1005
![](https://rs.olm.vn/images/avt/0.png?1311)
F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2-2/2010
=>F=2008/2010=1004/1005
![](https://rs.olm.vn/images/avt/0.png?1311)
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
100% giải đúng đầu tiên:
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 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 \(K=\frac{1004}{1005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
F=2 .(1/2-1/4+1/4-1/6+......+1/2008 - 1/2010)
= 2.(1/2-1/2010)
= 2. 502/1005
= 1004/1005
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)
=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
=\(2\left(1-\frac{1}{1005}\right)\)
=\(2.\frac{1004}{1005}\)
=\(\frac{2008}{1005}\)
P/s: Không biết đúng không nữa, làm đại ^.^
![](https://rs.olm.vn/images/avt/0.png?1311)
F=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
chú ý : \ là ngoặc
Gọi A= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
A/2= 2/2.4+2/4.6+...+2/2008.2010
Mà 2/2.4=1/2-1/4; 2/4.6=1/4-1/6 ....
Vậy A/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
A/2=1/2-1/2010=2010/4020-2/4020=2008/4...
A= 2008.2/4020=1004/1005
C = 4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2008.2010
C = 2 . (2/2.4 + 2/4.6 + 2/6.8 + ... + 2/2008.2010)
C = 2 . (1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2008 - 1/2010)
C = 2 . (1/2 - 1/2010)
C = 2 . 502/1005
C = 1004/1005