mateforum.ro Forum Index mateforum.ro

 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Fractie reductibila ?

 
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a
View previous topic :: View next topic  
Author Message
alex2008
Leibniz


Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Sun Nov 09, 2008 8:47 am    Post subject: Fractie reductibila ? Reply with quote

Se da fractia \frac{7+5^{2n+1}\cdot2^{2n}}{4+5^{3n}\cdot2^{3n+1}}, cu n \in N. Stabiliti daca este reductiblia.
Back to top
View user's profile Send private message
Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Thu Apr 23, 2009 10:18 pm    Post subject: Reply with quote

Pentru n=0: \frac{7+5^{2n+1}\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n+1}}=\frac{7+5}{4+2}=\frac{12}{6}, fractie reductibila.
Pentru n\geq 1:\frac{7+5^{2n+1}\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n+1}}=\frac{7+5^{2n}\cdot 5\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n}\cdot 2}=\frac{7+5\cdot 10^{2n}}{4+2\cdot 10^{3n}}\frac{500.....007}{200......004}.
Dar 7+0+0+...+0+0+5=12\Rightarrow numaratorul \vdots\ 3\ (1).
Dar 2+0+0+...+0+0+4=6\Rightarrow numitorul \vdots\ 3\ (2).
Din (1) si (2)\Rightarrow fractia este reductibila.
Back to top
View user's profile Send private message Send e-mail Yahoo Messenger
Display posts from previous:   
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a All times are GMT + 2 Hours
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum



Powered by phpBB © 2001, 2005 phpBB Group