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> Ne Ide Pa To Ti Je, Ali stvarno

predrag007
post Feb 6 2012, 01:21 PM
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Zadatak ide ovako:
Ako je x+y+z=1,x^2+y^2+z^2=1 i x^3+y^3+z^3=1 dokazati da je xyz=0.
Hvala unapred.
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pyost
post Feb 6 2012, 02:20 PM
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QUOTE(predrag007 @ Feb 6 2012, 01:21 PM)
Zadatak ide ovako:
Ako je x+y+z=1,x^2+y^2+z^2=1 i x^3+y^3+z^3=1 dokazati da je xyz=0.
Hvala unapred.
*



x+y+z=1\ \ =>\ \ (x+y+z)^2 = 1

(x+y+z)^2 - (x^2+y^2+z^2) = 1-1

=>\ \  2xy+2xz+2yz = 0\ \ =>\ \  xy+xz+yz = 0



(x+y+z)^3 - (x^3+y^3+z^3) = 1-1

=>\ \ 3 x^2 y +3 xy^2 +3 x^2 z + 3 xz^2 + 3 y^2 z + 3 yz^2 + 6xyz = 0

=>\ \  x^2 y + xy^2 + x^2 z+xz^2 + y^2 z + yz^2 + 2xyz = 0\ \ \ (1)



(xy+xz+yz)(x+y+z) = 1 \cdot 0

=>\ \  x^2 y + xy^2 + x^2 z+xz^2 + y^2 z + yz^2 + 3xyz = 0\ \ \ (2)


Na kraju oduzmes (1) od (2) i dobijes resenje.


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predrag007
post Feb 6 2012, 04:16 PM
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Hvala puno!!!!!!!
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