如何解方程x²- x=1?
1个回答
展开全部
法一:x³-x
=x(x²-1)
=x(x+1)(x-1)
法二:x³-x
=x³-1-x+1
=(x³-1)-(x-1)
=(x-1)(x²+x+1)-(x-1)
=(x-1)(x²+x+1-1)
=(x-1)(x²+x)
=x(x-1)(x+1)
法三:x³-x
=x³+1-x-1
=(x³+1)-(x+1)
=(x+1)(x²-x+1)-(x+1)
=(x+1)(x²-x+1-1)
=(x+1)(x²-x)
=x(x-1)(x+1)
=x(x²-1)
=x(x+1)(x-1)
法二:x³-x
=x³-1-x+1
=(x³-1)-(x-1)
=(x-1)(x²+x+1)-(x-1)
=(x-1)(x²+x+1-1)
=(x-1)(x²+x)
=x(x-1)(x+1)
法三:x³-x
=x³+1-x-1
=(x³+1)-(x+1)
=(x+1)(x²-x+1)-(x+1)
=(x+1)(x²-x+1-1)
=(x+1)(x²-x)
=x(x-1)(x+1)
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