设函数f(x)=|x^2+2x-1|,若a

设函数f(x)=|x^2+2x-1|,若a
设函数f(x)=|x^2+2x-1|,若a

设函数f(x)=|x^2+2x-1|,若a
(-3,-1)
画f（x）图易知-3且a^2+2a-1=-(b^2+2b-1)
所以(a+1)^2+(b+1)^2=4
令a=2sint-1,b=2cost-1
则-3/4π+2kπ所以ab+a+b=(2sint-1)*(2cost-1)+2sint-1+2cost-1=2sin2t-1
由于-3/2π+4kπ<2t<-π+4kπ
所以-1所以-3

ab+a+b [-3,1]