已知集合M={x|x=3n,n∈Z},N={x|x=3n+1,n∈Z},P={x|x=3n-1,n∈Z},且a∈M,b∈N,c∈P,设d=a-b+c,则

A.d∈MB.d∈NC.d∈PD以上都不对答案是a=3nb=3k+1c=3m-1d=3n-3k+3m-2=3n-3k+3m-3+1=3(n-k+m-1)+1但我唔明既然设... A.d∈M B.d∈N C.d∈P D以上都不对
答案是a=3n b=3k+1 c=3m-1
d=3n-3k+3m-2 =3n-3k+3m-3+1= 3(n-k+m-1)+1
但我唔明既然设了a=3n,点解唔设b系3n+1,C系3n-1
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奇巧且随和灬才俊2880
2011-06-29
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因为虽然M N P中的n都属于z,但a b c是三个集合中任意的三项,a b c中的n不一定相同。比如:a=3,b=4,c=2可以,但a=3,b=7,c=5也可以。n的取值在不同集合间不受影响,可任意取。
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那为什么 3(n-k+m-1)+1属于d∈N ?
n-k+m-1可能是负数
追答
因为n k m可取任何整数,说以(n-k+m-1)的值可以是任何整数。如果设这个数为n,d就刚好符合集合N,所以d∈N
即使n-k+m-1是负数,也是符合的,Z表示整数,包括负数,所以集合N中包含有负数
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因为在本题中n的取值可以一样也可以不一样,他就像是函数中的x一样,只是代表一个变量而已,你要是都设成n的话,就表示他们三个中的自变量必须都只能去一样的值,故不够全面所以在综合写在一个式子离得话应分开写成不同的字母。
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那为什么 3(n-k+m-1)+1属于d∈N ?
n-k+m-1可能是负数
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令e=a-b=3n-1
则d=e+c=3n-2=3n+1
选 B
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