f(x)=(ax-1)/(x+1),f’(x)=[(ax-1)’(x+1)-(ax-1)(x+1)’]/(x+1)2,=[a(x+1)-(ax-1)]/(x+1)2=(a+1)/(x+1)2要使f(x)在（-∞，-1）上是减函数，只要在（-∞，-1）上f’(x)<0恒成立，即a+1<0,a<-1.
