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...