已知集合A={x/3/(x+2)≥1},B={x/k-4<x≤2k+1},如果A∩B=空集

解.3/(x+2)≥1
即3/(x+2)-1≥0
即[3-(x+2)]/(x+2)≥0
即(x-1)/(x+2)≤0
解得-2<x≤1
A={x|-2<x≤1}
∵B={x|k-4<x≤2k+1},A∩B=空集
∴k-4≥1或2k+1≤-2
解得k≥5或k≤-3/2
∴实数k的取值范围为(-∞,-3/2]U[5,+∞)

x/3/(x+2)-1>=0
(x-3(x+2))/(3(x+2))>=0
(x+3)(x+2)<=0
x/=-2
-3<=x<-2
1.B空集:k-4>=2k+1
k<=-5
2.B/=空集:k>-5
2k+1<-3
or k-4>=-2
k<-2,k>=2
k<=-5
k<=-5