4个回答
2020-02-24 · 知道合伙人教育行家
关注
展开全部
(x²-x+3)²-(x²-x+3)
=(x²-x+3)[(x²-x+3)-1]
=(x²-x+3)(x²-x+2)
=(x²-x+3)[(x²-x+3)-1]
=(x²-x+3)(x²-x+2)
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
平方差公式
令a^2=(x²-x+3)² ,b^2=1
则,原式
=(x²-x+3+1)(x²-x+3-1)
=(x²-x+4)(x²-x+2)
令a^2=(x²-x+3)² ,b^2=1
则,原式
=(x²-x+3+1)(x²-x+3-1)
=(x²-x+4)(x²-x+2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
2020-02-24
展开全部
(x²-x+3)²-(x²-x+3)
=(x²-x+3)*(x²-x+3)-(x²-x+3)
=(x²-x+3)*{(x²-x+3)-1}
=(x²-x+3)*(x²-x+2)
=(x²-x+3)*x²-(x²-x+3)*x+(x²-x+3)*2
=(x^4-x³+3x²)-(x³-x²+3x)+(2x²-2x+6)
=x^4-2x³+6x²-5x+6
=(x²-x+3)*(x²-x+3)-(x²-x+3)
=(x²-x+3)*{(x²-x+3)-1}
=(x²-x+3)*(x²-x+2)
=(x²-x+3)*x²-(x²-x+3)*x+(x²-x+3)*2
=(x^4-x³+3x²)-(x³-x²+3x)+(2x²-2x+6)
=x^4-2x³+6x²-5x+6
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询