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Đặt A = 1 / 3 + 1 / 15 + 1 / 35 + 1 / 63 + 1 / 99 + 1 / 143 + 1 / 195
A = 1 / 1 x 3 + 1 / 3 x 5 + 1 / 5 x 7 +1 / 7 x 9 + 1 / 9 x 11 + 1 / 11 x 13 + 1 / 13 x 15
A x 2 = 2 / 1 x 3 + 2 / 3 x 5 +2/ 5 x 7 + 2/ 7 x 9 + 2 / 9 x 11 + 2/ 11 x 13 +2 / 13 x 15
A x 2 = 1 / 1 - 1 / 3 + 1 / 3 - 1 /5 + 1 / 5 - 1 / 7 + 1 / 7 - 1 / 9 + 1 / 9 - 1 / 11 + 1 / 11 - 1 / 13 + 1 / 13 - 1 / 15
A x 2 = 1 / 1 - 1 / 15
A x 2 = 14 / 15
A = 7 / 15
=1/3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15
=1-1/15
=14/15
vậy đáp số là 14/15
![](https://rs.olm.vn/images/avt/0.png?1311)
Dấu \(.\)là dấu nhân
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{2}.\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
~ Ủng hộ nhé
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
Suy ra ; \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}=\frac{14}{15}\)
=> A = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/15+1/35+1/63+1/99+1/143+1/195+1/255+1/323
=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19
=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19
=> 2A = 1/3 - 1/19
=> 2A = 16/57 => A = 16/57 : 2 = 8/57
=>=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19
=>=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19
=> 2A = 1/3 - 1/19
=> 2A = 16/57 => A = 16/57 : 2 = 8/57
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 1/35 + 1/63 + 1/99 + 1/143 + 1/195 + 1/255
= 1/ 5 x 7 + 1 / 7 x 9 + 1 / 9 x 11 + 1 / 11 x 13 + 1 / 13 x 15 + 1 / 15 x 17
= 1/5 - 1/7 + 1/ 7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 + 1/15 - 1/17
= 1/5 - 1/17
= 17/85 - 5/85
=12/85
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
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Đặt phép tính cần tìm là A
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(2A=1-\dfrac{1}{13}\)
\(2A=\dfrac{12}{13}\)
\(A=\dfrac{6}{13}\)
\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
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a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
Giải:
Đặt A = 1/3+1/15+1/35+1/63+1/99+1/143+1/195
2A= 2/(1.3) + 2/(3.5) + 2/(5.7) + 2/(7.9)+2/(9.11) + 2/(11.13)+2/(13.15)
2A=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9...
2A=1/1-1/15=14/15
Vậy A=14/15 : 2 = 7/15
Nhấn đúng mk nha
Tran Dan
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+..+\frac{1}{143}+\frac{1}{195}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{13.15}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{13}-\frac{1}{15}\)
= \(1-\frac{1}{15}=\frac{14}{15}\)
tick đúng nha