Răspunsuri

2014-08-17T21:08:28+03:00


Uc[x(x+1)]=Uc(x²)+ Uc(x)    Uc(x²)=0 ptr. x=0 ⇒ Uc[x(x+1)]=0

                                           Uc(x²)=1 ptr.x=1 sau 9⇒  Uc[x(x+1)]=2 sau 0

                                           Uc(x²) =4 ptr.x=2 sau8⇒ Uc[x(x+1)]=6 sau 2

                                          Uc(x²)=9 ptr.x=3 sau 7     Uc[x(x+1)]=2 sau 6

                                         Uc(x²)=6 ptr.x=4 sau 6      Uc[x(x+1)]=0 sau2

                                         Uc(x²)=5 ptr.x=5               Uc[x(x+1)]=0