數學的難題
回答 (3)
2009-04-15 4:44 am
✔ 最佳答案
求(x+1)6 除 (x2+2x-1)之餘式。sol
x^2+2x-1=0 ==>x^2=1-2x
(x+1)^6
==>[(x+1)^2]^3
==>[x^2+2x+1]^3
==>[1-2x+2x+1]^3
==>2^3
==>8
2009-04-15 4:56 am
求(x+1)6 除 (x2+2x-1)之餘式。
= 求(x+1)6 除(x2+2x+1)-2之餘式。
= 求(x+1)6 除 (x+1)2 - 2之餘式。
設 (x+1)2=P,
= 求P3 除 (P - 2) 之餘式。
= 23
= 8
= 求(x+1)6 除(x2+2x+1)-2之餘式。
= 求(x+1)6 除 (x+1)2 - 2之餘式。
設 (x+1)2=P,
= 求P3 除 (P - 2) 之餘式。
= 23
= 8
2009-04-15 4:53 am
No matter what method to can think of, I would rather use binomial theorem to expand (x+1)^6 and divide it by (x^2 + 2x -1) by long division to get the remainder. Seems that this is the simpliest method.
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