求y=√(x²-2x-3)+log(2)(x+2)的定义域

求y=√(x²-2x-3)+log(2)(x+2)的定义域
求y=√(x²-2x-3)+log(2)(x+2)的定义域

求y=√(x²-2x-3)+log(2)(x+2)的定义域
由题意可得：
x^2-2x-3≥0,x+2>0
联立解得：x≥3或,-2

x²-2x-3>=0且x+2>0,
截得，x>3

y=√(x²-2x-3)+log(2)(x+2)
case 1:
x²-2x-3 >=0
(x-3)(x+1) >=0
x >=3 or x<=-1
case 2:
(x+2) >=1
x >=-1
case 1 and case 2
"x >=3 or x<=-1" and " x>=-1"
ie x>=3 or x=-1
定义域 x>=3 or x=-1