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N
22 tháng 6 2018
N=7/2(2/1.3+....+2/13.15)
N=7/2.(1/1-1/3+.....+1/13-1/15)
N=7/2.(1-1/15)
N=7/2.(14/15)
N=7.14/2.15
QN
5
VP
3
26 tháng 3 2017
\(D=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+\dfrac{7}{99}+\dfrac{7}{143}\)
\(=7.\left(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\right)\)
\(=7.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)
\(=7.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{7}{2}.\left(\dfrac{1}{5}-\dfrac{1}{13}\right)\)
\(=\dfrac{7}{2}.\dfrac{8}{65}=\dfrac{28}{65}\)
TM
26 tháng 3 2017
D = 7\(\left(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\right)\)
= 7\(\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)
= 7\(\left[\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right):2\right]\)
= 7\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\): 2
= 7\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\) : 2
= 7 . \(\dfrac{10}{39}\) : 2 = \(\dfrac{70}{78}\) = \(\dfrac{35}{39}\)
WH
29 tháng 3 2018
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
31 tháng 7 2015
A = 5/3.7 + 5/7.11 + 5/11.15 + ... + 5/35.39
A = 5/4 x ( 4/3.7 + 4/7.11 + 4/11.15 + ... + 4/35.39 )
A = 5/4 x ( 1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/35 - 1/39 )
A = 5/4 x ( 1/3 - 1/39 )
A = 5/4 x 4/13
A = 5/13
ND
1
31 tháng 7 2015
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{35\cdot39}\)
\(=\frac{5}{4}\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{35\cdot39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...-\frac{1}{35}+\frac{1}{35}-\frac{1}{39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{39}\right)=\frac{5}{4}\cdot\frac{38}{39}=\frac{95}{78}\)
22 tháng 3 2019
\(7.\left[\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right):2\right]\)
\(7.\left[\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right):2\right]\)
\(7.\left(\frac{1}{3}-\frac{1}{13}\right):2\)
\(7.\frac{10}{39}:2=\frac{35}{39}\)
22 tháng 3 2019
\(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+\frac{7}{99}+\frac{7}{143}\)
\(=\frac{7}{2}\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\frac{10}{39}\)
\(=\frac{35}{39}\)
T
22 tháng 7 2019
#)Giải :
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}=7\left[\frac{1}{7}\left(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\right)\right]\)
\(C=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\times\frac{3}{35}=\frac{21}{35}\)
=7(1/3+1/15+1/35+...+1/9999)
=7(1/1.3+1/3.5+1/5.7+...+1/99.101)
=7(1-1/3+1/5-1/7+1/7-...+1/99-1/101)
=7(1-1/101)
=7.100/101=700/101
kết quả mk trà lời đó sai òi chia 2 nửa bạn ạ =305/101