(x^6-2x^3+1)/(x^2-2x+1)用竖式或代定系数法无
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 10:26:57
(x^6-2x^3+1)/(x^2-2x+1)用竖式或代定系数法无
(x^6-2x^3+1)/(x^2-2x+1)用竖式或代定系数法
无
(x^6-2x^3+1)/(x^2-2x+1)用竖式或代定系数法无
(x^6-2x^3+1)/(x^2-2x+1)
=(x^4+2x^3+3x^2+2x+1)(x^2-2x+1)/(x^2-2x+1)
=x^4+2x^3+3x^2+2x+1
=(x²+x+1)²
用竖式除得,不方便用电脑打出
设(x^6-2x^3+1)/(x^2-2x+1)=(x^4+ax^3+bx^2+cx+1)
则 x^6-2x^3+1
= (x^2-2x+1)(x^4+ax^3+bx^2+cx+1)
=x^2(x^4+ax^3+bx^2+cx+1)-2x(x^4+ax^3+bx^2+cx+1)+(x^4+ax^3+bx^2+cx+1)
=x^6+ax^5+bx^4+cx^3+x^2...
全部展开
设(x^6-2x^3+1)/(x^2-2x+1)=(x^4+ax^3+bx^2+cx+1)
则 x^6-2x^3+1
= (x^2-2x+1)(x^4+ax^3+bx^2+cx+1)
=x^2(x^4+ax^3+bx^2+cx+1)-2x(x^4+ax^3+bx^2+cx+1)+(x^4+ax^3+bx^2+cx+1)
=x^6+ax^5+bx^4+cx^3+x^2-2(x^5+ax^4+bx^3+cx^2+x)+(x^4+ax^3+bx^2+cx+1)
然后合并同类项,与 x^6-2x^3+1比较
5,4,3,2,1次项的系数分别为:0,0,-2,0,0
解得a,b,c的值,过程略
收起
(x^3-1)^2/(x-1)^2
=((x-1)(x^2+x+1)/(x-1))^2
=(x^2+x+1)^2
希望对你有帮助