\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{2009x2011}\)
K
Khách
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Những câu hỏi liên quan
DP
11 tháng 7 2017
\(\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{2009\cdot2011}+\frac{1}{x}=\frac{1}{5}\cdot0,5\)
\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{2011-2009}{2009\cdot2011}+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2009}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left(\frac{1}{2}\cdot\frac{2006}{10055}\right)+\frac{1}{x}=\frac{1}{10}\)
\(=\frac{1003}{10055}+\frac{1}{x}=\frac{1}{10}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{10}-\frac{1003}{10055}\)
\(\frac{1}{x}=\frac{1}{4022}\)
\(\Rightarrow x=1\div\frac{1}{4022}=4022\)
LC
2
8 tháng 7 2016
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
IW
8 tháng 7 2016
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
M
12 tháng 8 2018
p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
KT
12 tháng 8 2018
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)
M
11
13 tháng 7 2015
\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{11}\)
\(S.2=\frac{10}{11}\)
\(S=\frac{10}{11}:2\)
\(S=\frac{5}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 9 2023
A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2008}{6033}\)
A = \(\dfrac{1004}{6033}\)
NM
16 tháng 9 2023
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)
Đến đây bn tự tính nhé.
HA
Tìm x:
\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{19x21}\right).x=\frac{9}{7}\)
1
ML
23 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)
\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)
\(\frac{1}{7}.x=\frac{9}{7}\)
\(x=\frac{9}{7}\div\frac{1}{7}\)
\(x=9\)
Vậy ...
IM
31 tháng 8 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
6 tháng 5 2018
Tớ không chép lại đề nữa nhé:
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)=\(\frac{1}{2}.\left(\frac{3-1}{1-3}+\frac{7-5}{5-7}+...+\frac{2011-2009}{2009-2011}\right)\)
= \(\frac{1}{2}.\left(\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{2011}{2009.2011}-\frac{2009}{2009.2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\frac{2010}{2011}\)
=\(\frac{1005}{2011}\)
18 tháng 5 2019
\(A=\frac{1}{3}-\frac{1}{17}=\frac{14}{51}\)
cách làm thì tự biết
trên mạng đầy
18 tháng 5 2019
kết quả đúng phải là 7/51 chứ bn
mk cần cách trình bày thôi
câu trả lời của bn hơi lạnh nhạt tí ^.^
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2008}{6033}=\frac{1004}{6033}\)
\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{2009x2011}\)
\(=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+....+\frac{1.2}{2009.2011.2}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2008}{6033}=\frac{2008}{12066}\)