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Fractii

 
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MariusG



Joined: 16 Mar 2008
Posts: 4
Location: Rm.Valcea, Romania

PostPosted: Tue Feb 17, 2009 12:02 pm    Post subject: Fractii Reply with quote

Se dau fractiile:
Code:

A=(2004a+2005c)/(2005b+2004c);
B=(2004b+2005c)/(2005c+2004a);
C=(2004c+2005b)/(2005c+2004b);

Determinati a,b,c in Z stiind ca A,B,C sunt simultan numere intregi
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Al3xx
Euclid


Joined: 07 Nov 2008
Posts: 39
Location: Slatina

PostPosted: Tue Feb 17, 2009 12:21 pm    Post subject: Reply with quote

O sa restructurez datele problemei

A=\frac{2004a+2005c}{2005b+2004c}

B=\frac{2004b+2005c}{2005c+2004a}

C=\frac{2004c+2005b}{2005c+2004b}

Cred ca iese destul de usor cu divizibiliate.
Cum A \in Z => (2005b+2004c) | (2004a+2005c)

B \in Z => (2005c+2004a) | (2004b+2005c)

C \in Z => (2005c+2004b) | (2004b+2005c)

Folosim proprietatea divizibilitatii a|a

 => (2005b+2004c) | (2005b+2004c)
 => (2005c+2004a) | (2005c+2004a)
 => (2005c+2004b) | (2005c+2004b)

Scadem relatiile s.a.m.d. si ar trebui sa rezulte niste solutii simetrice de tipul
a=1
b=1
c=1


Last edited by Al3xx on Tue Feb 17, 2009 12:23 pm; edited 1 time in total
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Virgil Nicula
Euler


Joined: 28 Sep 2007
Posts: 672

PostPosted: Tue Feb 17, 2009 12:23 pm    Post subject: Re: Fractii Reply with quote

MariusG wrote:
Se dau fractiile:
Code:

A=(2004a+2005c)/(2005b+2004c);
B=(2004b+2005c)/(2005c+2004a);
C=(2004c+2005b)/(2005c+2004b);

Determinati a,b,c in Z stiind ca A,B,C sunt simultan numere intregi


MariusG, invata si tu LaTeX - ul. Era chiar mai simplu de scris. Nu mai vorbesc de "frumos".

MariusG wrote:
Se dau fractiile A=\frac {2004a+2005c}{2005b+2004c}\ ,\ B=\frac {2004b+2005c}{2005c+2004a}\ ,\ C=\frac {2004c+2005b}{2005c+2004b}\ .

Determinati numerele intregi a\ ,\ b\ ,\ c stiind ca A ,\ B\ ,\ C sunt simultan numere intregi.
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MariusG



Joined: 16 Mar 2008
Posts: 4
Location: Rm.Valcea, Romania

PostPosted: Tue Feb 17, 2009 7:27 pm    Post subject: Reply with quote

Al3xx wrote:
O sa restructurez datele problemei

A=\frac{2004a+2005c}{2005b+2004c}

B=\frac{2004b+2005c}{2005c+2004a}

C=\frac{2004c+2005b}{2005c+2004b}

Cred ca iese destul de usor cu divizibiliate.
Cum A \in Z => (2005b+2004c) | (2004a+2005c)

B \in Z => (2005c+2004a) | (2004b+2005c)

C \in Z => (2005c+2004b) | (2004b+2005c)

Folosim proprietatea divizibilitatii a|a

 => (2005b+2004c) | (2005b+2004c)
 => (2005c+2004a) | (2005c+2004a)
 => (2005c+2004b) | (2005c+2004b)

Scadem relatiile s.a.m.d. si ar trebui sa rezulte niste solutii simetrice de tipul
a=1
b=1
c=1


Mersi dar am incercat si eu dar nu a mers.
E undeva o lista cu comenzi LaTeX?
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