5已知集合M={x├|x=m+1/6 "," m"∈" Z}┤,N={x├|x=n/2 "-" 1/3 "," n"∈" Z}┤,则集合M,N的关系是(　　)

A.M⊆N B.M⫋N

C.N⊆M D.N⫋M

解析设n=2m或n=2m+1,m∈Z,则有

　　N={x├|x=2m/2 "-" 1/3 "或" x=(2m+1)/2 "-" 1/3 "," m"∈" Z}┤

　　={x├|x=m"-" 1/3┤┤或x=m+├ 1/6 "," m"∈" Z}.

　　又因为M={x├|x=m+1/6 "," m"∈" Z}┤,所以M⫋N.

答案B

6若非空数集A={x|2a+1≤x≤3a-5},B={x|3≤x≤22},则能使A⊆B成立的所有实数a的取值集合是0(　　)

A.{a|1≤a≤9} B.{a|6≤a≤9}

C.{a|a≤9} D.⌀

解析∵A为非空数集,∴2a+1≤3a-5,即a≥6.

　　又∵A⊆B,∴{■(2a+1≥3"," @3a"-" 5≤22"," )┤即{■(a≥1"," @a≤9"," )┤∴1≤a≤9.

　　综上可知,实数a的取值集合是{a|6≤a≤9}.

答案B

7已知集合A={1,3,6},集合B={3,a-2},若B⊆A,则实数a的值为　　　　　.

解析依题意,得a-2=1或a-2=6,解得a=3或a=8.

答案3或8

8已知A={a,0,-1},B={c+b"," 1/(a+b) "," 1},若A=B,则a=　　　　　,b=　　　　　,c=　　　　　.

解析由A=B,可知b+c=0,a=1,1/(a+b)=-1,

　　解得a=1,b=-2,c=2.

答案1　-2　2

9已知集合P={1,2,3,4},Q={0,2,4,5},则满足A⊆P,且A⊆Q的集合A为　　　　　.

解析若A=⌀,则满足A⊆P且A⊆Q;

　　若A≠⌀,由A⊆P且A⊆Q知集合A是由属于P且属于Q的元素构成,此时A可以为{2},{4},{2,4},故满足条件的集合A为⌀,{2},{4},{2,4}.

答案⌀,{2},{4},{2,4}

10已知集合A={x|x2-5x+6=0},B={x|(m-1)·x-1=0},且B⊆A,则以实数m为元素所构成的集合M为　　　　　.

解析A={x|x2-5x+6=0}={2,3}.

　　因为B⊆A,所以B=⌀或{2}或{3}.

当B=⌀时,⌀⊆A,满足题意,则m-1=0,即m=1;