tính
1/1 x3 + 1/3x5 + 1/5.7 +.........+1/ 2009x2011
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Những câu hỏi liên quan
2 tháng 2 2015
Câu 2: Bằng 50/101
(Đặt biểu thức đó là A. Nhân 2 vào 2A = 2/1x3+2/3x5+2/5x7+...+2/99x101
=>2A= 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101=100/101
=>A=100/101:2=50/101)
NB
1
23 tháng 4 2015
\(D=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{29.31}\)
\(\Rightarrow D=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+......+\frac{1}{2}.\left(\frac{1}{29}-\frac{1}{31}\right)\)
\(\Rightarrow D=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{31}\right)\)
\(\Rightarrow D=\frac{1}{2}.\frac{28}{93}=\frac{14}{94}\)
NHỚ LIKE CHO MÌNH NHA ^_^
5 tháng 4 2018
Ta có: \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)
\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}\)
\(=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
5 tháng 4 2018
N = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2005 - 1/2006
= 1/1 - 1/2006
= 2006/2006 - 1/2006
= 2005/2006
15 tháng 4 2021
Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)
\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)
\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)
\(\Leftrightarrow A=\dfrac{1006}{6045}\)
TT
15 tháng 4 2021
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)
2A=\(\dfrac{2014}{2015}\)
A=\(\dfrac{1007}{2015}\)
Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.
CC
2
13 tháng 4 2021
S=1/20+1/44+1/77+...+1/3080
S=1/20+1/44+1/77+...+1/3080
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=1/5−1/8+1/8−1/11+1/11−1/14+...+1/77−1/80
S.3/2=1/5-1/8+1/8-1/11+1/11-1/14+...+1/77-1/80
S.3/2=1/5−1/80
S.3/2=1/5-1/80
S.3/2=3/16
S.3/2=3/16
S=3/16:3/2
S=3/16:3/2
S=1/8
XO
16 tháng 8 2019
2) \(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{99\times101}+\frac{3}{101\times103}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{101\times103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\frac{101}{103}\)
\(=\frac{303}{206}\)
25 tháng 4 2016
vì 1/9 > 1/40 ; 1/29 > 1/40 ; 1/31 > 1/40; 1/39 > 1/40
nên 1/9 + 1/ 29 + 1/31 + 1/39 > 1/40 + 1/40 + 1/40 + 1/40 mà 1/40 + 1/40 + 1/40 + 1/40 = 1/10
=) M > 1/10
TB
25 tháng 4 2016
M > 1/20 + 1/30 + 1/40 + 1/40
M> 2/15 > 2/20 = 1/10
=> M > 1/10
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{2011-2009}{2009\cdot2011}\)
\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2009\cdot2011}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\frac{2010}{2011}=\frac{1005}{2011}\)
Đặt \(A=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2009\times2011}\)
\(2A=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2009\times2011}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(2A=1-\frac{1}{2011}=\frac{2010}{2011}\Rightarrow A=\frac{1005}{2011}\)