- 集合与命题(A,B)
- 已知集合A=｛x｜x-a＞0｝,集合B=｛x｜(x^2-2ax-3(a^2))/(x^2-x+1)＜0｝,求A∩B与A∪B
- 1.对于集合B=｛x｜(x^2-2ax-3(a^2))/(x^2-x+1)＜0}
∵(x^2-x+1)=(x-1/2)^2+3/4>0
∴由(x^2-2ax-3(a^2))/(x^2-x+1)＜0,得
x^2-2ax-3(a^2)<0,即(x+a)(x-3a)<0,
(1)当a>0,-aa
由上述两条,得出如下结论:
(1)当a>0
A∩B={x｜a-a)
(2)当a<0
A∩B={x｜a3a}
(3)当a=0
A∩B=空集
A∪B=A