S=2/5.7+2/7.9+2/9.11+.........+2/93.95
K
Khách
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TP
2
YY
24 tháng 4 2018
\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+..............+\frac{2}{93\cdot95}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+............+\frac{1}{93}-\frac{1}{95}\)
\(=\frac{1}{5}-\frac{1}{95}\)
24 tháng 4 2018
S = 2/5.7+ 2/7.9+...+2/93.95
=1/5-1/7+1/7-1/9+1/11+...+1/93+1/95
=1/5-1/95
=19/95-1/95
=18/95
NM
1
20 tháng 4 2015
S=\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+....+\(\frac{2}{93.95}\)+\(\frac{3}{95.98}\)+\(\frac{4}{98.102}\)+\(\frac{5}{102.107}\)+\(\frac{2012}{107.2119}\)
S=\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+....+\(\frac{1}{93}\)-\(\frac{1}{95}\)+\(\frac{1}{95}\)-\(\frac{1}{98}\)+\(\frac{1}{98}\)-\(\frac{1}{102}\)+\(\frac{1}{102}\)-\(\frac{1}{107}\)+\(\frac{1}{107}\)-\(\frac{1}{2119}\)
S=\(\frac{1}{5}\)-\(\frac{1}{2119}\)
S=\(\frac{2114}{10595}\)
21 tháng 5 2017
Khoảng cách có rồi thì bạn áp dụng công thức : \(\frac{a}{m.n}=\frac{1}{m}-\frac{1}{n}\)(với n-m=a) là làm được
N
21 tháng 5 2017
S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{93.95}+\frac{3}{95.98}+\frac{4}{98.102}+\frac{5}{102.17}+\frac{2012}{107..2119}\)
S=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+\frac{1}{102}-\frac{1}{107}+\frac{1}{107}-\frac{1}{2119}\)
S=\(\frac{1}{5}-\frac{1}{2119}\)
S=\(\frac{2114}{10595}\)
ND
10 tháng 5 2022
\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)
GM
1
3 tháng 4 2017
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
TT
3
TN
6 tháng 6 2016
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
6 tháng 6 2016
\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
Số dư dài quá. Đến đây bạn tự làm tiếp nhé
Bg
Ta có: S = \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
=> S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
=> S = \(\frac{1}{5}-\frac{1}{95}\)
=> S = \(\frac{19}{95}-\frac{1}{95}\)
=> S = \(\frac{18}{95}\)
Vậy S = \(\frac{18}{95}\)
\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{93\cdot95}\)
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(S=\left(\frac{1}{5}-\frac{1}{95}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{93}+\frac{1}{93}\right)\)
\(S=\left(\frac{1}{5}-\frac{1}{95}\right)\)
\(S=\frac{19}{95}-\frac{1}{95}\)
\(S=\frac{18}{95}\)