Question

y=lg(x^2-x-6)的定义域为A,y=根号(x+2)/(x+1)的定义域为B,求A∩B

要有过程
根号包住整个(x+2)/(x+1)

Answer 1

x^2-x-6>0
x>3 or x<-2 A
(x+2)/(x+1)>=0
x>-1 or x<=-2 B
A∩B为x<-2 or x>3

Answer 2

A的定义域：X>3或X<-2
B的定义域：X>-1或X<-2
答案：X>3或<-2