Točno
14. travnja 2012. 12:44 (12 godine, 3 mjeseci)
Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
Kliknite ovdje kako biste prikazali rješenje.
imamo a, d=a+1, P=a+2, S=a+3;
kako bi P i S po vieteovim formulama bili cijeli brojevim vrijedi
b=am
c=an
sad uvrstimo
a, a^2(m^2-4n), n, -m
a=a, n=a+2, -m=a+3 uvrstimo u a^2(m^2-4n)
dobimo:
(a^3+a^2-1)(a+1)=0
1. a=-1
2. a^3+a^2-1=0; a^3+a^2=-1; a^2(a+1)=1; tj.:
a+1=+-1 a=0; a=-2 ako uvrstimo u a^2(a+1)=1 vidjet cemo da ova jednadžba nema cjelobrojnih rješenja.
znači, a=-1
n=1
-m=2; m=-2
d=a^2(4-4)=0
-1,0,1,2; i to su jedina rješenja
kako bi P i S po vieteovim formulama bili cijeli brojevim vrijedi
b=am
c=an
sad uvrstimo
a, a^2(m^2-4n), n, -m
a=a, n=a+2, -m=a+3 uvrstimo u a^2(m^2-4n)
dobimo:
(a^3+a^2-1)(a+1)=0
1. a=-1
2. a^3+a^2-1=0; a^3+a^2=-1; a^2(a+1)=1; tj.:
a+1=+-1 a=0; a=-2 ako uvrstimo u a^2(a+1)=1 vidjet cemo da ova jednadžba nema cjelobrojnih rješenja.
znači, a=-1
n=1
-m=2; m=-2
d=a^2(4-4)=0
-1,0,1,2; i to su jedina rješenja
Ocjene: (2)
Komentari:
grga, 14. travnja 2012. 17:55
Filip_Wee, 14. travnja 2012. 16:48
grga, 14. travnja 2012. 14:34
grga, 14. travnja 2012. 14:33
treba ti bit
cini mi se.
i.. kuzis da uvodis oznake
, i ![-m = S](/media/m/4/a/9/4a960d4d35104b4dc0dae73356a87ea8.png)
mozda bi bilo dobro da si na drzavnom napisat da se radi od jednadbi
![-x^2 + 2x - 1](/media/m/d/9/2/d92a0c990a678695f48b79fe20eb7181.png)
pa onda jos provjerit kao dal je to zadovoljeno, za svaki slucaj
osim toga je tocno! :)
ponovo ti saljem tvoje rjesenje u latexu.
imamo
,
,
,
;
kako bi P i S po vieteovim formulama bili cijeli brojevim vrijedi
![b=am](/media/m/2/f/f/2fff471935b2803a098611b1903f88f9.png)
![c=an](/media/m/0/9/9/099be2e35f830ab7add5ad64ca1576f4.png)
sad uvrstimo
,
,
, ![-m](/media/m/c/3/c/c3ce2821eb28931b11617c1293cd0399.png)
,
,
uvrstimo u ![a^2(m^2-4n)](/media/m/8/8/d/88dfae10a7456e744e0c9a89749aa391.png)
dobimo:
![(a^3+a^2-1)(a+1)=0](/media/m/4/4/b/44b346e8ee277bfd16a45b1b3a77b2aa.png)
1.![a=-1](/media/m/e/4/b/e4b01fa69afcff262de3effc8186eee5.png)
2.
;
;
; tj.:
;
ako uvrstimo u
vidjet cemo da ova jednadžba nema cjelobrojnih rješenja.
znači,![a=-1](/media/m/e/4/b/e4b01fa69afcff262de3effc8186eee5.png)
![n=1](/media/m/4/e/4/4e466fe58c2a8f6389234c5c673f069c.png)
; ![m=-2](/media/m/f/9/d/f9dac033879f88ad2d9c3194709a13e0.png)
![d=a^2(4-4)=0](/media/m/7/1/d/71d09f0632e741f1a1d8248954cf854e.png)
,
,
,
; i to su jedina rješenja
![b = -am](/media/m/d/d/1/dd15bc1a698a569ca8102542ec2daaac.png)
i.. kuzis da uvodis oznake
![n = P](/media/m/2/8/2/2826312f3684424d396cafd6d6068d3f.png)
![-m = S](/media/m/4/a/9/4a960d4d35104b4dc0dae73356a87ea8.png)
mozda bi bilo dobro da si na drzavnom napisat da se radi od jednadbi
![-x^2 + 2x - 1](/media/m/d/9/2/d92a0c990a678695f48b79fe20eb7181.png)
pa onda jos provjerit kao dal je to zadovoljeno, za svaki slucaj
osim toga je tocno! :)
ponovo ti saljem tvoje rjesenje u latexu.
imamo
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
![d=a+1](/media/m/2/9/f/29f0e3ba849f62c325b2e4d87883a6b2.png)
![P=a+2](/media/m/9/0/0/9007390f800ae864ab2c21b6e2071c6d.png)
![S=a+3](/media/m/2/e/c/2ecae6cfc1e46c71689129d63ab02cd9.png)
kako bi P i S po vieteovim formulama bili cijeli brojevim vrijedi
![b=am](/media/m/2/f/f/2fff471935b2803a098611b1903f88f9.png)
![c=an](/media/m/0/9/9/099be2e35f830ab7add5ad64ca1576f4.png)
sad uvrstimo
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
![a^2(m^2-4n)](/media/m/8/8/d/88dfae10a7456e744e0c9a89749aa391.png)
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![-m](/media/m/c/3/c/c3ce2821eb28931b11617c1293cd0399.png)
![a=a](/media/m/e/2/5/e25307a3105249cdcef880149cb48082.png)
![n=a+2](/media/m/8/3/5/835ab7c87b147bfbeb90ff6caec98936.png)
![-m=a+3](/media/m/8/8/2/882c93569618acdd99c196a7c07be7f9.png)
![a^2(m^2-4n)](/media/m/8/8/d/88dfae10a7456e744e0c9a89749aa391.png)
dobimo:
![(a^3+a^2-1)(a+1)=0](/media/m/4/4/b/44b346e8ee277bfd16a45b1b3a77b2aa.png)
1.
![a=-1](/media/m/e/4/b/e4b01fa69afcff262de3effc8186eee5.png)
2.
![a^3+a^2-1=0](/media/m/9/9/9/99951957a977ae8c3d2d4ab314987211.png)
![a^3+a^2=-1](/media/m/c/7/b/c7b79b8d250933f3ba30d30bdba6238a.png)
![a^2(a+1)=1](/media/m/0/1/a/01aa64489ab3bf4dc73bd6256d2d04cb.png)
![a+1=+-1](/media/m/0/9/3/093d86ec2040f51ba448cc91d54570cf.png)
![a=0](/media/m/1/6/f/16ff59abcd4e49b52a5c57287e4536ee.png)
![a=-2](/media/m/1/a/a/1aa8852dab06ee4b68168322cd77f740.png)
![a^2(a+1)=1](/media/m/0/1/a/01aa64489ab3bf4dc73bd6256d2d04cb.png)
znači,
![a=-1](/media/m/e/4/b/e4b01fa69afcff262de3effc8186eee5.png)
![n=1](/media/m/4/e/4/4e466fe58c2a8f6389234c5c673f069c.png)
![-m=2](/media/m/5/4/a/54aae0063a8fcb488e40f9fc259eb024.png)
![m=-2](/media/m/f/9/d/f9dac033879f88ad2d9c3194709a13e0.png)
![d=a^2(4-4)=0](/media/m/7/1/d/71d09f0632e741f1a1d8248954cf854e.png)
![-1](/media/m/6/1/c/61cf05f5d8d6a4f0d373e7452cde9c3c.png)
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
![2](/media/m/e/e/e/eeef773d19a3b3f7bdf4c64f501e0291.png)