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Invers proportionalitate.

 
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Virgil Nicula
Euler


Joined: 28 Sep 2007
Posts: 672

PostPosted: Sun Jul 12, 2009 4:41 am    Post subject: Invers proportionalitate. Reply with quote

Se considera numerele x,y,z pentru care xyz\ne 0 si \frac 1x+\frac 1y+\frac 1z=\frac {1}{xyz} .

Sa se dea un exemplu de numere x,y,z care verifica relatiile date. Sa se arate

ca sirurile 1+x^2\ ,\ 1+y^2\ ,\ 1+z^2 si y+z\ ,\ z+x\ ,\ x+y sunt invers proportionale.
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sun Jul 12, 2009 11:32 am    Post subject: Reply with quote

x=1; y=\frac{1}{2}; z=\frac{1}{3}.
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\Rightarrow  xy+yz+zx=1.
(1+x^2)(y+z)=y+z+x(xy+xz)=y+z+x(1-yz)=y+z+x-xyz.
(1+y^2)(z+x)=z+x+y(yz+yx)=z+x+y(1-zx)=z+x+y-zxy.
(1+z^2)(x+y)=x+y+z(zx+zy)=x+y+z(1-xy)=x+y+z-zyx.
Atunci, (1+x^2)(y+z)=(1+y^2)(z+x)=(1+z^2)(x+y)=x+y+z-xyz\Rightarrow sirurile 1+x^2,\ 1+y^2,\ 1+z^2 si  y+z,\ z+x,\ x+y sunt invers proportionale.
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