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> Mgf Mathletics, NERDZ only! XD

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Wolfbrother
post Mar 7 2008, 08:53 PM
Post #101





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Woohoo


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"It's a known fact... math is the spawn of satan. Algebra? SATANIC! Pre-cal? SATANIC! Calculus? SATANIC! it is all horrible!"
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Wolfbrother
post Mar 7 2008, 09:02 PM
Post #102





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              WOLFBROTHER                says:
moram sad da nadjem
              WOLFBROTHER                says:
zadatak
              WOLFBROTHER                says:
za mgfmathwtfbbq
     *unicef Stefan_chemist     says:
mathletics XD.gif
     *unicef Stefan_chemist     says:
kao atletika
     *unicef Stefan_chemist     says:
samo je ovo matletika XD.gif


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Slovenatz
post Mar 7 2008, 09:06 PM
Post #103





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lol.gif misterija resena


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Wolfbrother
post Mar 7 2008, 09:24 PM
Post #104





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Ako je x,\ y,\ z\ \geq\ 0 i x\ +\ y\ +\ z\ =\ 4^{\frac{1}{3}} dokazi:

x^3\ +\ y^3\ +\ z^3\ +\ \frac{15xyz}{4}\ \geq\ 1

Ne mogu da nadjem treci koren kako se pise:(
Ajte ova je laka nejednakost(pazite kad sam je ja malo pojacao XD.gif)


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Nostradamus
post Mar 8 2008, 03:37 AM
Post #105





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Iz Šurove nejednakosti imamo

x^3 + y^3 + z^3 + 3xzy \geq x^2y + x^2z + z^2y + z^2x + y^2x + y^2z. Pomnozimo tu nejednakost sa tri. Sada pocetnu nejednakost pomnozimo sa 4 i dobija se
4(x^3 + y^3 + z^3) + 15xyz =

= 3x^3 + 3y^3 + 3z^3 + 9xzy + x^3 + y^3 + z^3 + 6xzy \geq

\geq x^3 + y^3 + z^3 + 6xzy + 3x^2y + 3x^2z + 3z^2y + 3z^2x + 3y^2x + 3y^2z

= (x + y + z)^3 = 4

I to je to

Napomena od: pyost
Razbijajte jednacine u vise redova, lakse je za citanje i ne remeti izgled foruma. Takodje, za razmak u TEXu se moze koristiti "\ " (kosa crta i razmak).


This post has been edited by pyost: Mar 14 2008, 12:01 PM


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 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 8 2008, 07:27 AM
Post #106





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Naravno

Vidim odsad moram postavljati teze zadatke tongue.gif


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Nostradamus
post Mar 8 2008, 03:46 PM
Post #107





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E nego, sta si mislio pod time da si pojacao nejednakost?


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 8 2008, 04:35 PM
Post #108





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Bila je nesto laksa i nije dostizala jednakost pa sam je malo izmenio, ali je u svakom slucaju prelaka. Ali sam za sledeci put nasao jednu lepu i malo tezu.

This post has been edited by Wolfbrother: Mar 8 2008, 04:35 PM


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Nostradamus
post Mar 8 2008, 07:13 PM
Post #109





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QUOTE(Wolfbrother @ Mar 8 2008, 04:35 PM)
Bila je nesto laksa i nije dostizala jednakost pa sam je malo izmenio, ali je u svakom slucaju prelaka. Ali sam za sledeci put nasao jednu lepu i malo tezu.
*


Aj stavi tu onda, nekako nemam ideja za zadatak


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 8 2008, 09:58 PM
Post #110





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Ako su x i y pozitivni realni brojevi dokazi:

 x^y+y^x>1


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Wolfbrother
post Mar 14 2008, 08:57 AM
Post #111





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Ima iko barem ideju? blink.gif

Imate tri slucaja x,y>1, x>1 y<1 x,y<1
Prva dva su trivijalna, ali treci je zeznut


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Nostradamus
post Mar 14 2008, 11:53 AM
Post #112





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Pa teoretski bi moglo da se uradi preko izvoda, ali ne verujem da se tako radi, a osim toga nemam ideje.

I ne, nisam izvalio ta 3 slucaja


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 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 14 2008, 02:49 PM
Post #113





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Pa ja bogami nisam uspeo da resim ovaj zadatak bez malo virenja u resenje sad.gif XD.gif


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Wolfbrother
post Mar 18 2008, 03:03 PM
Post #114





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Hoces da postujem resenje?


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username
post Mar 18 2008, 03:26 PM
Post #115





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sto se mene tice moze
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Wolfbrother
post Mar 18 2008, 03:44 PM
Post #116





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Wolfbrother
post Mar 18 2008, 03:45 PM
Post #117





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nostradamuse ti si ionako bio na redu da postavis zadatak.


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Nostradamus
post Mar 18 2008, 06:48 PM
Post #118





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Elementaran zadatak, nema sta, prosto ne znam kako nisam umeo da ga resim. biggrin.gif Inace ja sam mislio da moze da se uradi preko jensenove no nije sad ni bitno. Lep zadatak sve u svemu...

Funkcija f: R - R, i vazi x + f(x)=f(f(x)) za svako x iz R. Naci sva resenja f(f(x))=0


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 19 2008, 05:12 PM
Post #119





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Cek ja treba da nadjem sve x za koje je f(f(x))=0?


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Nostradamus
post Mar 19 2008, 08:08 PM
Post #120





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da


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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