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DAU COROANA
Se considera functia f:R–>R f(x)=3x-2.Rezolvati ecuatia f(x+1)+ f(x-2)=f(x)+4

Răspuns :

MFM
f(x)=3x-2
f(x+1)+ f(x-2)=f(x)+4
3x+3-2+3x-6-2=3x-2+4
6x-7=3x+2
3x=9
x=3

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