y=x^2-8x+15/x^2-x-6=[(x-3)(x-5)]/[(x-3)(x+2)]=(x-5)/(x+2)=(x+2-7)/(x+2)=1-7/(x+2) x≠3,x≠-2.因为7/(x+2)≠0,所以1-7/(x+2) ≠1因为x≠3,所以1-7/(x+2) ≠-2/5函数y=x^2-8x+15/x^2-x-6的值域是{y|y≠-2/5且y≠...