Tính tổng: S = \(\frac{1}{2011.2009}\)- \(\frac{1}{2009.2007}\)- ....... - \(\frac{1}{3.1}\)
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26 tháng 4 2017
\(=\frac{1}{2}.\left(\frac{1}{2011}-\frac{1}{2009}+\frac{1}{2009}-....+\frac{1}{3}-1\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2011}-1\right)\)
\(=\frac{1}{2}.\frac{-2012}{2011}=\frac{-1006}{2011}\)
TZ
1
6 tháng 8 2017
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{2}.\left(\frac{2}{99.97}-\frac{2}{97.95}-\frac{2}{95.93}-\frac{2}{5.3}-\frac{2}{3.1}\right)\)
\(=\frac{1}{2}.\left(\frac{99-97}{99.97}-\frac{97-95}{97.95}-\frac{95-93}{95.93}-\frac{5-3}{5.3}-\frac{3-1}{3.1}\right)\)
\(=\frac{1}{2}.\left[\left(\frac{99}{99.97}-\frac{97}{99.97}\right)-\left(\frac{97}{97.95}-\frac{95}{97.95}\right)-\left(\frac{95}{95.93}-\frac{93}{95.93}\right)-\left(\frac{5}{5.3}-\frac{3}{5.3}\right)-\left(\frac{3}{3.1}-\frac{1}{3.1}\right)\right]\)
\(=\frac{1}{2}.\left[\left(\frac{1}{97}-\frac{1}{99}\right)-\left(\frac{1}{95}-\frac{1}{97}\right)-\left(\frac{1}{93}-\frac{1}{95}\right)-\left(\frac{1}{3}-\frac{1}{5}\right)-\left(\frac{1}{1}-\frac{1}{3}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\frac{1}{95}+\frac{1}{97}-\frac{1}{93}+\frac{1}{95}-\frac{1}{3}+\frac{1}{5}-\frac{1}{1}+\frac{1}{3}\right]\)
\(=\frac{1}{2}.\left[-\frac{1}{99}-\frac{1}{93}+\frac{1}{5}-\frac{1}{1}\right]\)
DV
1
N
22 tháng 7 2017
\(\frac{1}{99.97}-\frac{1}{97.95}-........-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.........+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.......+\frac{1}{5.3}+\frac{1}{3.1}\right).\frac{2}{2}\)
\(=-\left(-\frac{2}{99.97}+\frac{2}{97.95}+......+\frac{2}{5.3}+\frac{2}{3.1}\right).\frac{1}{2}\)
\(=-\left(-\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{95}+.....+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}-1\right).\frac{1}{2}\)
\(=\left(\frac{1}{99}-1\right).\frac{1}{2}\)
\(=-\frac{98}{99}.\frac{1}{2}\)
\(=-\frac{49}{99}\)
LT
1
XO
23 tháng 8 2019
\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)
\(=\frac{1}{2}.\frac{2}{97.99}-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)
\(=\frac{1}{2}.\left[\frac{2}{97.99}-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}-\frac{98}{97}\right)\)
\(=\frac{1}{2}.\left(-1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{-100}{99}\)
\(=-\frac{50}{99}\)
13 tháng 8 2017
Tôi thấy bài này nó cứ sai sai
Ở chỗ \(\frac{1}{99.97}-\frac{1}{97.95}\)í
\(\frac{1}{97.95}>\frac{1}{99.97}\)mà ông Thám Tử THCS Nguyễn Hiếu CTV
13 tháng 8 2017
violympic cho sai đề :
Đề đúng là tính : \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.53}-....-\frac{1}{5.3}-\frac{1}{3.1}\)
Làm theo đề đúng !! ok
Ta có : \(A=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.53}+....+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)
VD
26 tháng 6 2016
đặt tổng là A
=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+....+\frac{1}{5}-\frac{1}{3}\right)\)
=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{3}\right)=\frac{32}{297}\)
LA
15 tháng 3 2017
a)\(\frac{1}{99.97}\)−\(\frac{1}{97.95}\)−\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)−\(\frac{1}{3.1}\)
=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(\(\frac{1}{95}\)−\(\frac{1}{97}\)+\(\frac{1}{93}\)−\(\frac{1}{95}\)+…+\(\frac{1}{3}\)−\(\frac{1}{5}\)+1−\(\frac{1}{3}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)
=\(\frac{-4751}{9603}\)
18 tháng 10 2016
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)