2. (x-1)/3=1/(y+1)⇒(x-1)(y+1)=3, x, y∈Z⇒ (x-1)∈Z, (y+1)∈Z
x-1=1, x=2
y+1=3, y=2 (2;2)
x-1=-1, x=0
y+1=-3, y=-4 (0;-4)
S={(0;-4); (2;2)}
3. abc:21 =c rest 12
abc numar in baza 10, a, b, c cifre, a≠0
abc=21q+12
dam valori lui q, pana gasim primul nr de 3 cifre
q=4, abc=84+12=96 nu are 3 cifre
q=5, abc=105+12=117
solutia este 117
5. (x²-x+1)²=(x²-x+1)(x²-x+1)=x^4-x³+x²-x³+x²-x+x²-x+1=x^4-2x³+3x²-2x+1
la 4 trebuie desen si nu am posibilitatea asta
