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Fie \( f:R->R \) astfel incat \( f(x^2-y^2)=(x+y)(f(x)-f(y)) \). Aratati ca \( f(2009x)=2009f(x) \).

Pt. \( x=y \)obtinem\( f(0)=0 \); pt. \( y=0 \), \( f(x^2)=xf(x)=-x(f-x) \), adica\( f(-x)=-f(x) \).
Luand \( y,-y \) avem \( f(x^2-(-y)^2)=(x-y)(f(x)+f(y)) => \)
\( (x+y)(f(x)-f(y))=(x-y)(f(x)+f(y)) => yf(x)=xf(y) => \frac{f(x)}{x}=\frac{f(y)}{y},\forall x,y \in \mathbb{R}. \)
Pt. \( y=1=>\frac{f(x)}{x}=f(1)=> f(x)=xf(1) => f(2009x)=2009xf(1)=2009(xf(1))=2009f(x), \) adica c.c.t.d.