题目内容
已知集合![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_ST/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_ST/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_ST/3.png)
A.M=N?P
B.M?N=P
C.M?N?P
D.N?P?M
【答案】分析:N={x|x=
,n∈Z},x=
=
,n∈Z;P={x|x=
,P∈Z},x=
=
;N=
=
=p,M={x|x=m+
,m∈Z},x=m+
=
,M,N,P三者分母相同,所以只需要比较他们的分子.M:6的倍数+1,N=P:3的倍数+1,所以M?N=P.
解答:解:N={x|x=
,n∈Z},
x=
=
,n∈Z.
P={x|x=
,P∈Z},
x=
=
,
N=
=
=p,
M={x|x=m+
,m∈Z},
x=m+
=
,
M,N,P三者分母相同,
所以只需要比较他们的分子.
M:6的倍数+1,
N=P:3的倍数+1,
所以M?N=P,
故选B.
点评:本题考查集合的包含关系的判断及其应用,是基础题.解题时要认真审题,仔细解答,注意合理地进行等价转化.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/9.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/10.png)
解答:解:N={x|x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/11.png)
x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/12.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/13.png)
P={x|x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/14.png)
x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/16.png)
N=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/18.png)
M={x|x=m+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/19.png)
x=m+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/20.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231137941826964/SYS201311012311379418269005_DA/21.png)
M,N,P三者分母相同,
所以只需要比较他们的分子.
M:6的倍数+1,
N=P:3的倍数+1,
所以M?N=P,
故选B.
点评:本题考查集合的包含关系的判断及其应用,是基础题.解题时要认真审题,仔细解答,注意合理地进行等价转化.
![](http://thumb.zyjl.cn/images/loading.gif)
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