设集合A={x|log1/2(6+x-x^2)>-2},B={x|根号下ax-a^2|

Question

设集合A={x|log1/2(6+x-x^2)>-2},B={x|根号下ax-a^2|

Answer 1

A:log1/2(6+x-x^2)>-2=log1/2(4),
6+x-x^2<4,(x-2)(x+1)>0,
x>2或x<-1
B:-x+3a<√ax-a^2x>(a^2+3a)/(√a+1)且(√a-1)x1)a=1,0<-2,错,B为空集,取
2)0<=a<1,
x>(a^2-3a)/(√a-1)=a(a-3)(√a+1)/(a-1)=a(a√a+a-3√a-3)/(a-1),
且x>(a^2+3a)/(√a+1)=a(a+3)(√a-1)/(a-1)=a(a√a-a+3√a-3)/(a-1)
a(a√a+a-3√a-3)/(a-1)-a(a√a-a+3√a-3)/(a-1)
=2a√a(√a-3)/(a-1)>0
所以x>a(a-3)(√a+1)/(a-1)
与A相交不可能为空,故舍.（其实根本就无需讨论谁大谁小,因为皆为“>”,好傻啊~）
3)a>1,(a^2+3a)/(√a+1)1,B为空集
(a^2+3a)/(√a+1)>=(a^2-3a)/(√a-1)
解得a<=9,即12,B不为空集
-1<=(a^2+3a)/(√a+1)<(a^2-3a)/(√a-1)<=2
即a^2+3a+√a+1>=0,恒成立,a>1,
且a>9
且a^2-3a-2√a+2<=0
把a^2-3a-2√a+2<=0,解了,就OK啦,
你自己看着办吧 
可证明a>9时,a^2-3a-2√a+2<=0恒不成立（发个图给你看看就知道了）,即B不为空集时,a无解
综上所述：1<=a<=9 
你这题很花时间哎~,别忘了加分,如果正确的话