∵x²=1, y²=4, xy=2
∴（x²-1)(y²-1)=0, 展开：x²y²-y²-x²+1=4-y²-x²+1=5-y²-x²=0

y²+x²-5=0
xy=2亲，xy=2是二元二次方程。(x-1)(y-1)=0(x-1)(y-1)=xy-y-x+1=2-y-x+1=3-x-y=0,即x+y-3=0(x+1)(y+1)=xy+y+x+1=2+y+x+1=3+x+y=0,即x+y+3=0 y²+x²-5=0x+y-3=0或y²+x²-5=0x+y+3=0