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AK
1 tháng 6 2018
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
1 tháng 6 2018
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 = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)
NN
18 tháng 8 2017
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
NH
18 tháng 8 2017
(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
N
25 tháng 7 2018
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
25 tháng 7 2018
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, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
MP
25 tháng 1 2016
A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
DD
5 tháng 1 2016
\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)
A.2=2/3.5+2/5.7+2/7.9+…+2/99.101
A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101
Vậy
A.2=1/3-1/101=98/303
A=98/303/2=49/303
Đúng không
4 tháng 1 2016
A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101
= 1/3 - 1/101 = 98/303
Vậy A = 98/303 : 2 = 49/303
2 tháng 2 2016
1/
A= 1/15+1/35+1/63+1/99+ ... + 1/9999
A=1/3.5+1/5.7+1/7.9+ ... +1/99.101
2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101
2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101
2A=1/3-1/101
A=49/303
Sai thì thôi nhé
2 tháng 2 2016
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
14 tháng 4 2019
\(=1-\frac{1}{3}+\frac{2}{3.5}+...+\frac{2}{15.17}\)
\(=1-\frac{1}{3}+2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{15.17}\right)\)
\(=1-\frac{1}{3}+2.\left(\frac{1}{3}-\frac{1}{17}\right)\)
\(=\frac{62}{51}\)
14 tháng 4 2019
\(\Leftrightarrow A=1-\frac{1}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{13.15}+\frac{2}{15.17}\)
\(\Leftrightarrow A=\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{13.15}+\frac{2}{15.17}\)
\(\Leftrightarrow A=1-\frac{1}{17}=\frac{16}{17}\)
27 tháng 6 2017
= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101
=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)
=1/2 x (1/3 - 1/99)
=1/2 x (1/3 - 1/101)
=1/2 x (98/303)
=1/15 + 1/35 + 1/63 +1/99+...+1/9999
=49/303
27 tháng 6 2017
\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)
\(=\frac{98}{303}\)
N
18 tháng 7 2017
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
TN
18 tháng 7 2017
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)=\frac{1}{2}.\frac{14}{15}\)\(=\frac{7}{15}\)
b)\(\frac{1414+1515+...+1919}{2020+2121+...+2525}\)
\(\Rightarrow\frac{101\left(14+15+16+17+18+19\right)}{101\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}\)
\(=\frac{\left(19+14\right).6:2}{\left(25+20\right).6:2}=\frac{19+14}{25+20}=\frac{33}{45}=\frac{11}{15}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\)
\(\Rightarrow S=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(\Rightarrow S=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+....+\frac{1}{195}\)
\(=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{13x15}\)
\(=\frac{1}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{13x15}\right)\)
\(=\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}x\left(1-\frac{1}{15}\right)=\frac{1}{2}x\frac{14}{15}=\frac{7}{15}\)