百威英博官网:X^10+X^5+1=?

来源：百度文库编辑：神马品牌网时间：2024/04/27 17:39:41

因式分解

x^10+x^5+1
=(x-1)(x^10+x^5+1)/(x-1)
=(x^11-x^10+x^6-x^5+x-1)/(x-1)
其中x^11-x^10+x^6-x^5+x-1
=(x^11-x^5)-(x^10-x)+(x^6-1)
=(x^6-1)(x^5+1)-x(x^9-1)
=(x^3-1)(x^3+1)(x^5+1)-x(x^3-1)(x^6+x^3+1)
=(x^3-1)[(x^8+x^5+x^3+1)-(x^7+x^4+x)]
=(x^3-1)(x^8-x^7+x^5-x^4+x^3-x+1)
∴原式=(x^3-1)(x^8-x^7+x^5-x^4+x^3-x+1)/(x-1)
=(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)

X^10+X^5+1＝X^10+X^5+1＋X^5－X^5
＝（X^5＋1）^2－X^5
＝(X^5＋1－X^2/5)(X^5＋1+X^2/5)

X^10+X^5+1=? x^10+x^5+1 x^6+x^5+x^4+x^3+x^2+x+1=0 5x-2/(x+2)(x-2)-(x-1)(x+2)/(x+2)(x-2)=0 f(x)=(x-1)(x-2)(x-3).......(x-10) 求f'(9)=? x*(x/3)=(x-5)*(x/3+1)/2 x=? (x-4)(x-2)-(x-1)(x+3),其中x=-2/5 已知f(x)=5-x+@(3x+1) (x^2-1)(x+3)(x=5)分解因式 X-28=(100-X)-[20- 5分之1X ]?