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2 tháng 2 2019
a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2007x2009}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2009}\right)=\frac{1}{2}\cdot\frac{2008}{2009}=\frac{1004}{2009}\)
....
các bài cn lại bn lm tương tự nha
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
b, \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
3A = \(\dfrac{1}{6}+\dfrac{1}{18}+...+\dfrac{1}{330}\)
3A-A = \(\dfrac{1}{6}-\dfrac{1}{990}\)
2A = 82/495
A =82/495 : 2
A=41/495
3 tháng 5 2015
K:2=2/2.4+2/4.6+2/6.8+...+2/2008.2010
=1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010
=1/2-1/2010
=502/1005
K=502/1005.2
=1004/1005
F=1/3.6+1/6.9+1/9.12+...+1/30.33
3F=3/3.6+3/6.9+3/9.12+...+1/30.33
=1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33
=1/3-1-33
=10/33
F=10/33:3
=10/99
7 tháng 3 2020
Bn sai câu K = 4/2.4 + 4/4.6 + 4/6.8 +....+ 4/2008.2010
10 tháng 4 2019
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)
7 tháng 5 2021
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)
\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)
MT
17 tháng 7 2015
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\left(\frac{1008}{2016}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)
K
17 tháng 7 2015
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{2014.2016}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}\)
\(=\frac{1007}{1008}\)
NV
3
2 tháng 10 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.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))
=> A = 2.\(\frac{502}{1005}\)
=> A = \(\frac{1004}{1005}\)
2 tháng 10 2016
đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\)
=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}\)
=\(\frac{502}{1005}\)
Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)
PT
9
4 tháng 8 2016
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{2014.2016}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}=\frac{1007}{1008}\)
LH
4 tháng 8 2016
\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}\)
\(A=\frac{4}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)
27 tháng 3 2017
\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)
\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)
\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10.\dfrac{49}{100}\)
\(\Rightarrow5B=\dfrac{49}{10}\)
Vậy \(5B=\dfrac{49}{10}\)
27 tháng 3 2017
Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).
=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).
=> B = \(\dfrac{49}{200}\).
=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).
Vậy 5B = \(\dfrac{49}{40}\).
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
\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)
\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)
\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)
K = đề bài
= 2 . ( 2/2.4 + 2/4.6 + 2/6.8 + . . . + 2/2008.2010 )
= 2 . ( 1 - 1/4 + 1/4 - 1/6 + 1/8 - 1/8 + . . . + 2/2008 - 2/2010 )
= 2 . ( 1 - 2/2010 )
= ( phần còn lại bạn tự tính nha )
k cho mình đó, bài này mình làm rồi nên đúng 100% lun, sorry nha mình ngại viết nhiều
a) \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Leftrightarrow K=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\frac{2009}{2010}=\frac{2009}{1005}\)
b) F=1/18 + 1/54 + 1/108 +...+ 1/990
=> \(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(\Leftrightarrow F=3.\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\right)\)
\(\Leftrightarrow F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(\Leftrightarrow F=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(\Leftrightarrow F=1-\frac{1}{33}=\frac{32}{33}\)