y=(x+1)/(x²+2x+2)的值域
求Δ法
(x²+2x+2)y=x+1
yx²+2yx+2y=x+1
yx²+(2y-1)x+(2y-1)=0
y≠0
且Δ=(2y-1)²-4y(2y-1)>=0
(2y-1)(-2y-1)>=0
(2y-1)(2y+1)<=0
-1/2<=y<=1/2,且y≠0
当y=0时
-x-1=0
x=-1
y=0成立
∴综上值域是[-1/2,1/2]