y=3x+2/5-4x 求定义域和值域

y=(3x+2)/(5-4x)
定义域,分母不为0
∴ 5-4x≠0
即 x≠5/4
即定义域是{x|x≠5/4}
值域
y=(3x+2)/(5-4x)
=[(-3/4)(5-4x)+2+15/4]/(5-4x)
=(-3/4)+(23/4)/(5-4x)
∵ 5-4x≠0
∴ (23/4)/(5-4x)≠0
∴ (-3/4)+(23/4)/(5-4x)≠-3/4
即值域是{y|y≠-3/4}