x²+x+1=(x+ 1/2)²+ 3/4
x²+x+1≠0
故定义域x∈R
原式整理得 x²(2-y)-x(1+y)-(1+y)=0
△=(1+y)²+4(2-y)(1+y)≥0
3y²-6y-9≤0
y²-2y-3≤0
(y-3)(y+1)≤0
解得-1≤y≤3
x²+x+1=(x+ 1/2)²+ 3/4
x²+x+1≠0
故定义域x∈R
原式整理得 x²(2-y)-x(1+y)-(1+y)=0
△=(1+y)²+4(2-y)(1+y)≥0
3y²-6y-9≤0
y²-2y-3≤0
(y-3)(y+1)≤0
解得-1≤y≤3