1/1.3+1/3.5+1/5.7+......+1/99.101
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12 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
LT
Lương Thị Vân Anh
CTVHS
12 tháng 3 2023
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
TT
1
30 tháng 9 2015
6F=1.3(5-1)+3.5(7-1)+5.7(9-3)+...99.101(103-97)
6F=1.3+1.3.5-1.3.5+3.5.7-3.5.7+.....-97.99.101+99.101.103
6F=3+99.101.103
6F=3+1029897
6F=1029900
F =1029900:6
F=171650
NT
3
4 tháng 5 2018
=1/2(2/1.3+2/3.5+2/5.7+....+2/2009.2011
=1/2(1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-1/2011
=1/2(1/1-1/2011)
=1/2.2010/2011
=1005/2011
4 tháng 5 2018
=1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-2011
=1-1/2011
=2010/1011
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.
TM
1
16 tháng 9 2020
Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow\frac{x+2}{2021}=1\)
\(\Leftrightarrow x=2019\)
Vậy \(x=2019\)
TM
0
16 tháng 7 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
16 tháng 7 2016
Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9
2A=2/1.3+2/3.5+2/5.7+2/7.9
2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
2A=1-1/9
2A=8/9
A=4/9
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{99\cdot101}\)
\(=2\cdot\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2} \cdot\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}\cdot\frac{100}{101}\)
\(=\frac{50}{101}\)