tích trên có chữ số tận cùng là 5

neu nhu 1 so nhan voi 1;10;10;1000;...

thi so do cung co so tan cung la 0 nen chu so tan cung cua day :

1*3*5*7...*2009*2011 la 0

đầu bài là ê các bạn mình thấy lạ thật đấy chưa từng thấy đầu bài nào kì cục như vậy

= ( 1 + 3 + 5 + ... + 2011 ) - ( 2 + 6 + 8 + ... + 2010 )

= [ 1006 x ( 2011 + 1 ) : 2 ] - [ 1005 x ( 2010 + 2 ) : 2 ]

= ( 1006 x 2012 : 2 ) - ( 1005 x 2012 : 2 )

= ( 2024072 : 2 ) - ( 2022060 : 2 )

= 1012036 - 1011030

= 1006

m giong ban kia !ban nao tk mk di !

`#P.Vy`

`C=(3/3+2/3)(5/5+2/5)(7/7+2/7)...(2009/2009+2/2009)(2011/2011+2/2011)`

`C=5/3xx7/5xx9/7xx...xx2011/2009xx2013/2011`

\(C=\dfrac{5.7.9\times...\times2011.2013}{3.5.7\times...\times2009.2011}\)

`C=671`

https://olm.vn/hoi-dap/detail/105210184248.html

\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2009.2011}\)

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009.2011}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=2\left(1-\frac{1}{2011}\right)\)

\(=2.\frac{2010}{2011}\)

\(=\frac{4020}{2011}\)

Có:

A=4/1x3+4/3x5+4/5x7+....+4/2009x2011

1/2 x A=1/2x(4/1x3+4/3x5+4/5x7+....+4/2009x2011)

1/2 x A=2/1x3 + 2/3x5 + 2/5x7 +...+ 2/2009x2011

1/2 x A=1-1/3+1/3-1/5+1/5-1/7+.....+1/2009-1/2011

1/2 x A=1 - 1/2011

1/2 x A= 2010/2011

A = 4020/2011

Ủng hộ nha ko ủng hộ thì k

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, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)

= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

1/1x2 + 1/2x3+ 1/3x4+....+1/2009x2010

= 1/1-1/2 + 1/2-1/3+ 1/3-1/4+...+1/2009-1/2010

= 1/1-1/2010

= 2009/2010

\(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{2009\cdot2010}\)

\(=1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2010}\)

\(=1-\frac{1}{2010}\)

\(=\frac{2009}{2010}\)