C = 4/2.4+4/4.6+4/6.8+ ...... + 4/2008.2010
K=4/2.4+4/4.6+4/6.8+...+4/2008.2010
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
F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
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
F = 4/2.4 + 4/4.6 + 4/6.8 + ....+ 4/2008.2010
Gọi F= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
F/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 F/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
F/2=1/2-1/2010=2010/4020-2/4020=2008/4...
F= 2008.2/4020=1004/1005
Tính A=4/2.4+4/4.6+4/6.8+....+4/2008.2010
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: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
tính
K= 4/2.4 + 4/4.6 + 4/6.8+...+ 4/2008.2010
=> 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 \(K=\frac{1004}{1005}\)
Tính:
F= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
F=2 .(1/2-1/4+1/4-1/6+......+1/2008 - 1/2010)
= 2.(1/2-1/2010)
= 2. 502/1005
= 1004/1005
Tính nhanh: F= \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
Cảm ơn!
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
=2(2/2.4+2/4.6+......+22/008.2010)
=2(12−12010)
C=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{2008.2010}\)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)
=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
=\(2\left(1-\frac{1}{1005}\right)\)
=\(2.\frac{1004}{1005}\)
=\(\frac{2008}{1005}\)
P/s: Không biết đúng không nữa, làm đại ^.^
Ta thấy : \(\frac{4}{2.4}=\frac{1}{2}\left(\frac{4}{2}-\frac{4}{4}\right);\frac{4}{4.6}=\frac{1}{2}\left(\frac{4}{4}-\frac{4}{6}\right);...;\frac{4}{2008.2010}=\frac{1}{2}\left(\frac{4}{2008}-\frac{4}{2010}\right)\)
=> C =\(\frac{1}{2}.\left(\frac{4}{2}-\frac{4}{4}+\frac{4}{4}-\frac{4}{6}+\frac{4}{6}-\frac{4}{8}+...+\frac{4}{2008}-\frac{4}{2010}\right)\)
=> C = \(\frac{1}{2}\left(\frac{4}{2}-\frac{4}{2010}\right)=\frac{1}{2}\left(2-\frac{2}{1005}\right)=\frac{1}{2}\left(\frac{2010}{1005}-\frac{2}{1005}\right)\)
=> C = \(\frac{1}{2}\left(\frac{2010-2}{1005}\right)=\frac{1}{2}.\frac{2008}{1005}=\frac{1004}{1005}\)
Tính giá trị của biểu thức sau
F= 4/2.4+4/4.6+4/6.8 +...+4/2008.2010
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