yx²+xy+y=x²-x+1
(y-1)x²+(y+1)x+(y-1)=0
x是实数则方程有解
所以判别式大于等于0
所以(y+1)²-4(y-1)²>=0
(y+1+2y-2)(y+1-2y+2)>=0
(3y-1)(y-3)