楼上定义域都不讨论,有问题第一,y的值域先要讨论y=(1/2)+(3/2)/(4x+3)因为(3/2)/(4x+3)≠0所以y≠1/2y-1/2=(3/2)/(4x+3)4x+3=(3/2)/(y-1/2)4x+3=3/(2y-1)x=[3/(2y-1)-3]/4=3(1-y)/[2(2y-1)]y的范围是(-∞,1/2)∪(1/2...