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PT
2
27 tháng 4 2016
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11.13}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(A=1-\frac{1}{13}\)
\(A=\frac{12}{13}\)
27 tháng 4 2016
A = 1/3 + 1/15 + 1/35 + 1/63 +.....+ 1/143
= 1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 +.....+1/11.13
= 1/2 . (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +...+ 1/11 - 1/13)
= 1/2 . (1 - 1/13)
= 1/2 . 12/13
= 6/13
LT
3
11 tháng 4 2016
A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15
A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15
A=1 - 1/15
A=1/14
NH
3
15 tháng 7 2015
Gọi dãy là A ta có :
A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13
A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )
A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )
A = 1/2 . ( 1/3 - 1/13 )
A = 1/2 . 10/39
A = 5/39
HT
15 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
=\(\frac{1}{2}.\frac{10}{39}\)
=\(\frac{5}{39}\)
NH
2
15 tháng 7 2015
1/15+1/35+1/63+1/99+1/143
=1/2x(1/15+1/35+1/63+1/99+1/143)
=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)
=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
DP
3
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
25 tháng 7 2019
Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
\(A=\frac{5}{39}\)
Câu còn lại cx dựa như vậy nhé bn !
Chúc bn hc tốt <3
NT
0
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
NN
2
S
6 tháng 4 2018
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{9.11}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(tắt 1 bước nha)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{9}-\frac{1}{11}\)
\(2A=\frac{1}{3}-\frac{1}{11}\)
\(2A=\frac{8}{33}\)
\(\Rightarrow A=\frac{4}{33}\)
Vậy A=_____________
PV
11 tháng 4 2016
a) A x 3/5 = 3/1.4 + 3/4.7 + 3/7.10 + ...+ 3/100.103
= 1 - 1/4 + 1/4 - 1/7 + 1/7 -1/10 + ... + 1/100 - 1/103
= 1 - 1/103
= 102/103 => A = 102/103 : 3/5 = 170/103
b) B = 1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/49.51
B x2 = 2/3.5 + 2/5.7 + 2/7.9 + ...+ 2/49.51
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1.49 - 1/51
= 1/3 - 1/51
= 16/51
=> B = 16/51 : 2 = 8/51
\(A=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{3843}\)
\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{61\cdot63}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{61\cdot63}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{63}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{63}\right)=\dfrac{1}{2}\cdot\dfrac{20}{63}=\dfrac{10}{63}\)
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