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O inegalitate tot de clasa a cincea

 
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alex2008
Leibniz


Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Wed Nov 12, 2008 9:27 pm    Post subject: O inegalitate tot de clasa a cincea Reply with quote

Demonstrati ca \frac{a+2^{180}}{a+3^{120}}<\frac{b+5^{900}}{b+7^{720}}, oricare ar fi a si b numere naturale.
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Andi Brojbeanu
Newton


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

PostPosted: Mon Apr 20, 2009 5:26 pm    Post subject: Reply with quote

2^{180}=(2^3)^{60}=9^{60}; 3^{120}=(3^2)^{60}=8^{60};
5^{900}=(5^5)^{180}=3125^{180}; 7^{720}=(7^4)^{180}=2401^{180}.
Inmultind mezii cu extremii, obtinem a\cdot b+a\cdot 2401^{180}+8^{60}\cdot b+8^{60}\cdot 2401^{180}<a\cdot b+a\cdot 3125^{180}+9^{60}\cdot b+9^{60}\cdot 3125^{180}
sau 0<a(3125^{180}-2401^{180})+b(9^{60}-8^{60})+9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}(1).
Dar, 3125^{180}>2401^{180}\Rightarrow a(3125^{180}-2401^{180})\geq 0(2) si 9^{60}>8^{60}\Rightarrow b(9^{60}-8^{60})\geq 0(3).
De asemenea, inmultind cele doua relatii, obtinem 9^{60}\cdot 3125^{180}>8^{60}\cdot 2401^{180}\Rightarrow 9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}>0(4).
Adunand relatiile (2), (3) si (4) obtinem relatia (1), care este adevarata, ceea ce trebuia demonstrat.
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Laurian Filip
Site Admin


Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Mon Apr 20, 2009 5:41 pm    Post subject: Reply with quote

Prima fractie e subunitara, iar a doua e supraunitara. Cred ca e mai simplu de rezolvat asa. Very Happy
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