Решите систему уравнений:
{x^2+y^2-xy=3
{
{x+y-3xy=-3

{x^2+y^2-xy=3 {x+y-3xy=-3{(x+y)²-3xy=3{(x+y)-3xy=-3-----------------(x+y)²-(x+y)=6(x+y)²-(x+y)-6=0D=25=5²1)x+y=(1+5)/2=3y=3-xx+3-x-3(3-x)x=-33-9x+3x²+3=03x²-9x+6=0x²-3x+2=0D=1x₁=(3-1)/2=1   y₁=2x₂=(3+1)/2=2y₂=12)x+y=(1-5)/2=-2y=-x-2x+y-3xy=-3x-x-2-3x(-x-2)=-3-2+3x²+6x+3=03x²+6x+1=0D=36-12=24=4*6x₃=(-6-2√6)/6=-1-√6 /3=-1-√(2/3)y₃=-x-2=1+√(2/3)-2=√(2/3)-1x₄=-1+√(2/3)y₄=1-√(2/3)-2=-1-√(2/3)