x+1<0则f(x+1)=-(x+1)+1=-x所以x-x(x+1)<=1x²+1>=0恒成立则x<-1x+1>=0f(x)=(x+1)-1=x所以x+x(x+1)<=1x²+2x-1<=0-2-√2<=x<=-2+√2所以-1<=x<=-2+√2综上x≤-2+√2
