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erictan2046

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            Çór(-22)=?


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           ³ýÒÔ(x^2+x+1)(x+1)ʱµÄÓàʽ¡£

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math2000

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¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
erictan2046: ½ð±Ò+5, ¡ï¡ï¡ïºÜÓаïÖú 2012-08-11 21:08:35
ÎÊÌâ1
Éèr(x) = ax^2+bx+c
Ôò
f(x)=x^2345+x^23+45=x(x-1)(x+1)g(x)+r(x)
Áîx=0,1,-1´úÈËÉÏʽ£¬µÃµ½Èý¸ö·½³Ì£¬½â³öa/b/c¼´¿É
f(0)=45=r(0)=c
f(1)=47=r(1)=a+b+c
f(-1)=43=r(-1)=a-b+c
Óɴ˵õ½r(x)µÄ±í´ïʽ¡£
ÎÊÌâ2ÀàËÆÎÊÌâ1£¬×Ô¼º½â°É
2Â¥2012-08-11 19:22:10
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

erictan2046

Í­³æ (ÕýʽдÊÖ)

ÒýÓûØÌû:
2Â¥: Originally posted by math2000 at 2012-08-11 19:22:10
ÎÊÌâ1
Éèr(x) = ax^2+bx+c
Ôò
f(x)=x^2345+x^23+45=x(x-1)(x+1)g(x)+r(x)
Áîx=0,1,-1´úÈËÉÏʽ£¬µÃµ½Èý¸ö·½³Ì£¬½â³öa/b/c¼´¿É
f(0)=45=r(0)=c
f(1)=47=r(1)=a+b+c
f(-1)=43=r(-1)=a-b+c
Óɴ˵õ½r(x)µÄ±í´ïʽ ...

3Â¥2012-08-11 21:09:45
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

maqiaoyun

½ð³æ (СÓÐÃûÆø)

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¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
ÈôÒ»¸ö¶àÏîʽf(x)=x^2345+x^23+45±»£¨x^3-x£©³ýËùµÃÓàʽÊÇr(x),
            Çór(-22)=?
³ýÁËÀûÓÃÊÖ¹¤ÑÝËãµÃµ½a,b.cÍ⣬Ҳ¿ÉÖ±½ÓʹÓÃÊýѧÈí¼þ£¬ÈçmathemticaÖÐÊäÈëPolynomialRemainder[x^2345 + x^23 + 45, x^3 - x, x]£¬¼´µÃÓàʽ2x+45,ÓÚÊÇr(-22)=1
4Â¥2012-08-14 08:48:23
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

maqiaoyun

½ð³æ (СÓÐÃûÆø)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

¡ï ¡ï ¡ï
erictan2046: ½ð±Ò+3, ¡ï¡ï¡ïºÜÓаïÖú 2012-08-14 13:39:00
ÎÊÌâ2£ºµ±¶àÏîʽf(x)³ýÒÔx^2+x+1¼°£¨x+1£©^2ʱ£¬ÆäÓàʽ·Ö±ðΪx+5¼°x-1¡£Çóf(x) ³ýÒÔ(x^2+x+1)(x+1)ʱµÄÓàʽ¡£
½â£ºÒÀÌâÓÐf(x)=(x^2+x+1)q(x)+x+5
Éèq(x)=(x+1)p(x)+r,ÏÔÈ»r=q(-1)=f(-1)-4£¬´úÈëÉÏʽ
f(x)=(x^2+x+1)[(x+1)p(x)+f(-1)-4]+x+5
ÔÙÓÉf(x)=[(x+1)^2]g(x)+(x-1)Öªf(-1)=-2,»¯¼òÉÏʽµÃ
f(x)=(x^2+x+1)[(x+1)p(x)-6]+x+5
     =(x^2+x+1)(x+1)p(x)-6x^2-6x-6+x+5
     =(x^2+x+1)(x+1)p(x)-6x^2-5x-1
ËùÒÔ£¬ËùÇóÓàʽΪ-6x^2-5x-1
5Â¥2012-08-14 09:18:45
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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