a) 4x^2 -4x+1 -(x^2× 3+2xV3+1) = x^2-xV3-2x+2V3 -xV3-4x
4x^2 -4x+1 -3x^2-2xV3-1 = x^2-2xV3-6x
x^2 -x^2 -4x +6x -2xV3+2xV3 = 0
2x=0
x=0
b) V17 - 2V2 se poate scrie ca (3-2V2)^2
verificare (3-2V2)^2 = 9-8V2 +4×2 = 9-12V2+8= 17-12V2
[V(3-2V2)^2 + V(2V2+1)^2] / (x-1)
[3-2V2 + 2V2+1] / (x-1)
4/ x-1 apartine Z+
deci se impune ca x-1 sa fie printre divizorii lui 4
D4={-1, -2, -4, 2, 4}
x-1 = -1, x= 0 apartine Z+
x-1 = -2, x= -1 nu apartine Z+
x-1= -4, x= -3 nu apartine Z+
x-1= 1, x= 2 apartine Z+
x-1= 2, x= 3 apartine Z+
x-1= 4, x= 5 apartime Z+
deci A={0,2,3,5}
