NT
3
K
Khách
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LH
7
29 tháng 3 2017
B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13
= 1/2 * ( 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13)
= 1/2 * ( 1/3 - 1/5 + 1/5 -1/7 + ...+ 1/11 - 1/13)
= 1/2 * ( 1/3 - 1/11)
= 1/2 * 8/33
= 4/33
29 tháng 3 2017
\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(B=\frac{1}{2}.\frac{12}{39}\)
\(B=\frac{2}{13}\)
16 tháng 3 2016
1/15 + 1/35 + 1/63 + 1/99 + 1/143
đặt A = 1/15 + 1/35 + 1/63 + 1/99 + 1/143
A = 1/3X5 + 1/5X7 + 1/7X9 + 1/9X11 + 1/11X13
A x 2 = 2 x ( 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + 1/11x13 )
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13
A x 2 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13
A x 2 = 1/3 - 1/13
A x 2 = 13/39 - 3/39
A x 2 = 10/39
A =10/39 : 2
A = 5/39
TK
3
NT
0
LT
2
29 tháng 3 2016
=\(\frac{1}{3\cdot5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11\cdot13}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{13}=\frac{5}{13}\)
HM
29 tháng 3 2016
B=1/15+1/35+1/63+1/99+1/143
B=1/3.5+1/5.7+1/7.9+1/9.11+11.13 (khoảng cách từ 3-5;5-7;7-9;9-11;11-13 la 2)
Suy ra B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) (Ta gop -1/5+1/5;-1/7+1/7;-1/9+1/9;-1/11+1/11 bang 0)
B=1/2(1/3-1/43)=1/2.40/129=20/129
BT
3
18 tháng 4 2017
\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=> \(2B=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\) => \(2B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\) => \(2B=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\) => \(2B=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+\dfrac{11}{9.11}-\dfrac{9}{9.11}+\dfrac{13}{11.13}-\dfrac{11}{11.13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{13}\)
=> \(B=\left(\dfrac{13}{39}-\dfrac{3}{39}\right):2\)
=> \(B=\dfrac{10}{39}.\dfrac{1}{2}\)
=> \(B=\dfrac{10}{39.2}\)
=> \(B=\dfrac{5}{39}\)
Vậy \(B=\dfrac{5}{39}\)
AT
18 tháng 4 2017
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(B=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)
18 tháng 3 2016
\(B=\frac{7}{15}+\frac{4}{5}-1=\frac{19}{15}-1=\frac{4}{15}\)
\(5-\frac{2}{3}+\frac{1}{5}=\frac{13}{3}+\frac{1}{5}=\frac{68}{15}\)
PT
1
9 tháng 3 2016
Ta có:\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{5}{39}\)
B=1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + 1/11x13
B=1/2(2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13)
B=1/2(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13)
B=1/2(1/3 - 1/13)
B=1/2 x 10/39
B=5/39
B = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13
2B = 2 ( 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 )
2B = 1/3-1/5+1/5-1/7+1/7-1/9+.....+1/11-1/13
2B = 1/3 - 1/13
2B = 10/39
B = 10/39 . 1/2
B = 10/78