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BIG FOOT
Član broj: 2964
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*.ptt.yu
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Koliko ima celobrojnih resenja parametra k za  ako je x prirodan broj???
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| 28.11.2002. u 21:15 |
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sallle
Sasa Ninkovic
Beograd
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Poruke: 357
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ICQ: 20785904
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kx-kk=x+7
x(k-1)=kk+7
x=(kk+7)/(k-1) {pretpostavka da k nije 1}
x=(kk-1+8)/(k-1)
x=(kk-1)/(k-1)+8/(k-1)
x=k+1+8/(k-1)
8 mora da bude deljivo sa k-1, sto znaci:
k-1=+-1, k=2, k=0
k-1=+-2, k=3, k=-1
k-1=+-4, k=5, k=-3
k-1=+-8, k=9, k=-7
Za k=1 imamo (ovo gore resenje vazi ako k nije 1):
k=1, x-1=x+7 - jednacina nema resenja..
znaci ima 8 celobrojnih resenja...
poz,
sale
p.s. ebem li ga, nikako da naucim ovaj tex
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| 03.12.2002. u 00:21 |
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sallle
Sasa Ninkovic
Beograd
Član broj: 146
Poruke: 357
*.beg.sezampro.yu
ICQ: 20785904
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E zaebo sam se, posto videh sad da x mora biti prirodan broj, izbacuju se resenja:
k=0,k=-1,k=-3,k=-7
i ostaju:
k=2,k=3,k=5,k=9
poz
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| 03.12.2002. u 00:25 |
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