Find the remainder and quotient of [(x-699)^699 - (x-701)^701] / (2x - 1400)
註：請提供完整列式過程。
[] = 中括號
^ = 次方
F.4 Maths remainder & quotient
2012-04-09 1:27 am
回答 (3)
2012-04-09 4:09 am
✔ 最佳答案
以q及r 代表商及餘數:(x-699)^699-(x-701)^701=q(2x-1400)+R
以 700 代 X 就得到R.H.S.=R
1^699-(-1)^701=R<=由於701是單數,所以(-1)^701= -1
得到R=1-(-1)=2//
q= (x-699)^699-(x-701)^701-2
-------------------------------------
(2x-1400)
這是沒得化簡的死式黎- -可能我未學過掛....
2012-04-10 5:36 am
無錯，，，的確是我的失誤，抱歉
2012-04-09 21:38:06 補充：
[(x-699)^699 - (x-701)^701] - 2
= (x-699)^699 - 1 + (701 - x)^701 - 1
= (x-699-1)[(x-699)^698 +698(x-699)^697 + ... + (1)^698]
+ (701-x-1)[(701-x)^700 + ... + 1^700]
2012-04-09 21:38:06 補充：
[(x-699)^699 - (x-701)^701] - 2
= (x-699)^699 - 1 + (701 - x)^701 - 1
= (x-699-1)[(x-699)^698 +698(x-699)^697 + ... + (1)^698]
+ (701-x-1)[(701-x)^700 + ... + 1^700]
2012-04-10 12:11 am
...
= (x-699)^699 - 1 + (701 - x)^701 - 1
= (x-699-1)[x^698 +699x^697 + ... + (699)^698]
+ (701-x-1)[(701)^700 + ... + x^700]
= ...
好似有問題??
(x - 699)^699 - 1
= (x - 699 - 1)*[(x - 699)^698 + (x - 699)^697 + . . . + 1)
不應該是 (x-699-1)[x^698 +699x^697 + ... + (699)^698]
= (x-699)^699 - 1 + (701 - x)^701 - 1
= (x-699-1)[x^698 +699x^697 + ... + (699)^698]
+ (701-x-1)[(701)^700 + ... + x^700]
= ...
好似有問題??
(x - 699)^699 - 1
= (x - 699 - 1)*[(x - 699)^698 + (x - 699)^697 + . . . + 1)
不應該是 (x-699-1)[x^698 +699x^697 + ... + (699)^698]
收錄日期: 2021-04-13 18:36:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120408000051KK00579