[(1025^4+1025^2+1)/(1024^4+1024^2+1)]x[(1027^4+1027^2+1)/(1026^4+1026^2+1)]x.......x[(2047^4+2047^2+1)/(2046^4+2046^2+1)]之值最接近哪一個整數?

2016-11-01 9:51 am

回答 (1)

2016-11-01 10:31 am
✔ 最佳答案
Sol
x^4+x^2+1
=(x^4+2x^2+1)-x^2
=(x^2+1)^2-x^2
=(x^2+x+1)(x^2-x+1)
(x+1)^4+(x+1)^2+1
=[(x+1)^2+(x+1)+1]*[(x+1)^2-(x+1)+1]
=[(x+1)^2+(x+1)+1]*(x^2+x+1)
So
[(x+1)^4+(x+1)^2+1]/(x^4+x^2+1)=[(x+1)^2+(x+1)+1]/(x^2+x+1)
So
[(1025^4+1025^2+1)/(1024^4+1024^2+1)]*[(1027^4+1027^2+1)/(1026^4+1026^2+1)]*…*[(2047^4+2047^2+1)/(2046^4+2046^2+1)]
=[(2047^4+2047^2+1)/(2046^4+2046^2+1)]*[(2046^4+2046^2+1)/(2045^4+2045^2+1)]*…*[(1025^4+1025^2+1)/(1024^4+1024^2+1)]
=(2047^2+2047+1)/(1024^2+1024+1)
=4192257/1049601
=3.994
最接近整數4


收錄日期: 2021-04-11 21:32:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161101015131AABlI3G

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