Question

已知集合A={x^-ax+x-a>0}B={1/x-a-1≤-1},a属于R.(1)求A,B (2)

Answer 1

1.A={(x-a)(x+1)>0}
所以当a>-1时 A=｛x|x>a或x<-1｝
当a<-1时 A=｛x|x>-1或x<a｝
B={x-a/x-a-1≤0},
{（x-a)(x-a-1)≤0
x-a-1<>0
得B={a≤x<a+1},
2. -2<a<=-1

周末练习么？

Answer 2

(1)x^2-ax+x-a>0
(x-a)(x+1)>0
当a>-1时 A=｛x|x>a或x<-1｝
当a<-1时 A=｛x|x>-1或x<a｝
1/x-a-1<=-1
1/x-a<=0
1/x-ax/x<=0
(1-ax)/x<=0
若a>0 (ax-1)/x>=0 当a<0时 (-ax+1)x<=0
(ax-1)x>=0 B=｛x|-1/a<=x<=0｝
当a>1时 B=｛x|x>=0或x<=1/a｝
当0<a<1时 B=｛x|x>=a或x<=0｝