a=x²-3x²=a+3f(a)=lg(a+3)/(a+3-6)所以f(x)=lg[(x+3)/(x-3)]真数大于0(x+3)/(x-3)>0则(x+3)(x-3)>0x<-3x>3所以定义域(-∞,-3)∪(3,+∞)
(0,根号6)并上(-根号6,0)
(-根号6,正无穷)