用判别式法.
y(x²+x-1)=3x²+3x+1,整理得(y-3)x²+(y-3)x-y-1=0
若y=3,则无解
若y≠3,则△=(y-3)²+4(y+1)(y-3)≥0,解得y≥3或y≤-1/5
综上,y>3或y≤-1/5