Question

y=x^2-4x-5/x^2-3x-4值域

Answer 1

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}