M = 1/ 1.3 + 1/ 3.5 + 1/ 5.7 +.....+ 1/ 2003.2005
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11 tháng 7 2016
\(=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\frac{2004}{2005}\)
\(=\frac{1002}{2005}\)
28 tháng 3 2018
Đặt :
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{2003.2005}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+.......+\dfrac{2}{2003.2005}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.......+\dfrac{1}{2003}-\dfrac{1}{2005}\)
\(\Leftrightarrow2A=1-\dfrac{1}{2005}\)
\(\Leftrightarrow2A=\dfrac{2004}{2005}\)
\(\Leftrightarrow A=\dfrac{1002}{2005}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
7 tháng 7 2023
Đặt biểu thức là A
\(2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2005-2003}{2003.2005}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}:2=\dfrac{1002}{2005}\)
7 tháng 7 2023
Gọi tổng trên là A. Ta có
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{2005}=\dfrac{2005}{2005}-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
⇒ A= \(\dfrac{2004}{2005}:2=\dfrac{2004}{2005}.\dfrac{1}{2}=\dfrac{1002}{2005}\)
Vậy tổng trên bằng \(\dfrac{1002}{2005}\)
TN
1
9 tháng 1 2022
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2003\cdot2005}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2004}{2005}=\dfrac{1002}{2005}\)
PA
11 tháng 7 2016
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2001\times2003}+\frac{1}{2003\times2005}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2001\times2003}+\frac{2}{2003\times2005}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2001}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2005}\right)=\frac{1}{2}\times\left(1-\frac{1}{2005}\right)=\frac{1}{2}\times\frac{2004}{2005}=\frac{1002}{2005}\)
Chúc bạn học tốt
LB
0
CB
0
HN
5
19 tháng 3 2019
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2003.2005}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2003}-\frac{1}{2005}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2005}\right)=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)
19 tháng 3 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}=\)
\(=\frac{2}{2.1.3}+\frac{2}{2.3.5}+\frac{2}{2.5.7}+....+\frac{2}{2.2003.2005}\)
\(=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)
\(=\frac{1}{2}.\frac{2004}{2005}\)
\(=\frac{1002}{2005}\)
Chúc bạn học tốt nha!
J
1
S
7 tháng 8 2018
Đặt \(U=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
\(\Rightarrow U=\dfrac{1.2}{1.2.3}+\dfrac{1.2}{3.2.5}+\dfrac{1.2}{5.2.7}+...+\dfrac{1.2}{2003.2.2005}\)
\(\Rightarrow U=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{2005}\right)\Rightarrow U=\dfrac{1}{2}.\dfrac{2004}{2005}\Rightarrow U=\dfrac{1002}{2005}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2003.2005}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{2}.\dfrac{2004}{2005}=\dfrac{1002}{2005}\)
2M= 1/1.3+1/3.5+1/5.7+...+1/2003.2005
2M= 1/1-1/3+1/3-1/5+...+1/2003-1/2005
2M= 1/1-1/2005
2M= 2004/2005
M= 2004/2005:2
M=1002/2005