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Alta problema cu fractii

 
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crenguta.mihailescu



Joined: 18 Oct 2011
Posts: 3

PostPosted: Thu Nov 17, 2011 8:12 pm    Post subject: Alta problema cu fractii Reply with quote

Fie A= 2^2/1x3+4^2/3x5+6^2/6x7+...+2010^2/2009x2011

Aratati ca 1005<A<1005,5[/list]
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seby97
Euclid


Joined: 04 Aug 2011
Posts: 31

PostPosted: Fri Nov 18, 2011 6:22 pm    Post subject: Reply with quote

Problema aceasta a fost la concursul interjudetean Dimitrie Pompeiu clasa a 7a 2011
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