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,陵悔a b c中的n不一定相同。比告汪誉如:a=3,b=4,c=2可以,但a=3,b=7,c=5也可以。n的取值在不同集合袜段间不受影响,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
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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,只是代橡李表一个变量而已,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歼如睁式子离氏岁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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