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> Pismeni Iz Analize, trazim savet

username
post Nov 3 2007, 02:45 PM
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sutra imam pismeni iz analize, predaje mi mirjana perovanovic pa ako ima neko neki savet ili nesto sto treba da znam neka kaze
ne trebaju mi saveti o prepisivanju i puskicam necu da varam sleep.gif
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RZA
post Nov 5 2007, 10:46 PM
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Njeno Ljubičanstvo


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ah
pa kod perovanovicke moras da radis.
mislim.. na casu radi olakse zadatke.. al na kontrolnom ume bogami da zada neke teze smile.gif

saveta nemam.. sem.. radi..

ona je poprilicno pravedna..


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Gaara
post Nov 5 2007, 10:49 PM
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To uopste nije tacno...
Daje zadatke slicne kao i na casovima...
Ne brini se ako prodjes na republicko imas kod nje 5 sigurno... smile.gif (al ozbiljno)
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Slovenatz
post Nov 5 2007, 10:52 PM
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jes malo sutra, ja sam imao sve petice u drugom polugodistu u 8, nagradu na republickom i zakljucila mi je 4 na kraju, zbog 3+ u prvom dry.gif . Ali pismeni su laki, ne bi trebalo da imas nekih ozbiljnijih problema.


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Kuzman
post Nov 5 2007, 11:39 PM
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Mirjana je car!
Na casu se rade laki zadaci, a onda na kont/pismenom dodje neki spejs satl...


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username
post Nov 6 2007, 08:22 AM
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bio je pismeni tezi zadaci
jedan* je bio nezavistan od pismenog i ko ga resi dobije pet (ko sve resi dve petice smile.gif )
dobicu 3, 4 ili 5

*naci bijekciju koja preslikava A u B
A={xeR I 1<=x<=2}
B={xeR I -1<=x<=3}
ja sam odvalio f(x)=-2/3*x biggrin.gif

This post has been edited by username: Nov 6 2007, 08:40 AM
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username
post Nov 8 2007, 11:14 PM
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dobio sam 3 sad.gif
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username
post Nov 25 2007, 07:06 PM
Post #8





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da li neko moze da mi pomogne oko matematicke indukcije
1/(1*3) + 1/(3*5) + ... + 1/(2n-1)(2n+1) = n/(2n+1)
problem je kod 3. koraka:
n(2n+3)/(2n+1)(2n+3) = (n+1)/(2n+3)
i da li znate neki sajt sa zadacima iz ove oblasti?

This post has been edited by username: Nov 25 2007, 07:08 PM
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Nostradamus
post Nov 25 2007, 07:34 PM
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Nije bas preko indukcije, ali evo:
napisi 1/3*1 kao (3-1)/2*3*1, 1/5*3 kao (5-3)/2*3*5 i td.
svi ce imati 1/2 pa je izvuces i dobija se:

1/2((3-1)/3*1 + (5-3)/5*3 + ... + ((2n+1) - (2n-1))/(2n+1)*(2n-1))

e sad kad razvijes te razlomke dobijas

1/2(1 - 1/3 + 1/3 - 1/5 + 1/5 - ... -1/(2n+1)) t.j skrate se svi osim prvog i poslednjeg

1/2(1 - 1/(2n+1))=1/2(2n/(2n+1))=n/(2n+1).

Ako nesto nisi shvatio pitaj

Inace preko indukcije bi samo trebao da pokazes

n/(2n+1) + 1/(2n+1)*(2n+3)=(n+1)/(2n+3) =

(n(2n+3) + 1)/(2n+1)*(2n+3)=(n+1)/(2n+3) tj (pomnozis sa 2n+1)
n(2n+3) + 1 = (n+1)(2n+1)
2n*n + 3n + 1 = 2n*n + 2n + n + 1 i to je to


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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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username
post Nov 25 2007, 07:47 PM
Post #10





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da, nisam znao kako da rastavljam sve to
hvala
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username
post Feb 10 2008, 09:20 AM
Post #11





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kako se resava ovaj tip zadataka?

P(x+3)=x^2+2x+2
P(x)=?

This post has been edited by username: Feb 10 2008, 09:21 AM
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Hannibal Lecter
post Feb 10 2008, 11:47 AM
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Smena

t=x+3\ =>\ x=t-3

P(t)=(t-3)^2+2(t-3)+2

P(x)=(x-3)^2+2(x-3)+2


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Svi me žele, a ja sam nedodirljiv!
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username
post Feb 10 2008, 11:54 AM
Post #13





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zanimljivo smile.gif
ja sam nasao malo drugaciji nacin:
P(x+3)=x^2+2x+2

P(x)=ax^2+bx+c

a(x+3)^2+b(x+3)+c=x^2+2x+2

ax^2+(6a+b )x+9a+3b+c=x^2+2x+2

a=1

6a+b=2

=>b=-4

9a+3b+c=2

=>c=5

P(x)=x^2-4x+5

ali ovo sa smenom izgleda lakse i bolje

This post has been edited by username: Feb 10 2008, 11:57 AM
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pyost
post Feb 10 2008, 12:00 PM
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Deus Ex Makina
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Vazno je da si se ti snasao XD.gif


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Wolfbrother
post Feb 11 2008, 02:02 PM
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