y=x^2-4x-5/x^2-3x-4值域

y=x^2-4x-5/x^2-3x-4
=[(x+1)(x-5)]/[(x+1)(x-4)]
则[(x+1)(x-4)]不等于0
则x+1不等于0且x-4不等于0
x不等于-1且x不等于4
又y=[(x+1)(x-5)]/[(x+1)(x-4)]
=(x-5)/(x-4)
=[(x-4)-1]/(x-4)
=1-[1/(x-4)]
由x不等于4,则1/(x-4)不等于0
则y=1-[1/(x-4)]不等于1
又x不等于-1 则代入得y不等于6/5
则值域为{y|y不等于6/5且y不等于1}