1·3 +a·1- 2·3 +b= 3· 1+a·3 - 2·1 + b =1
a- 3 +b= 3· a -2 +b =1
a+b -3=1
3a+ b +1=1 ⇒ a+b=4
3a+b=0 la scadem ⇒ 2a= -4 a=-2 si b=6
b. folosim asociativitatea x ×( y×z) = ( x×y) ×z
egalam coeficientii pentru x :
y :
z :
c.rescriem legea x×y = (x -2 ) · ( y -2) +2
x×x= (x-2)·(x-2) +2= (x-2)² +2
[ x×x]×x = [ ( x-2)² +2 ] × x =[ ( x-2)² +2 -2 ]( x-2) +2 = (x-2) ³ +2
(x-2)³ +2 =3
( x-2) ³ - 1 =0 folosim formula diferenta cuburi
(x - 2-1 ) [ (x-2)² + (x-2) · 1 + 1² ]=0
x-3 =0 ⇒ x=3
paranteza dreapta , rezolvata are radacini complexe