(1+1/x)/(2x-1+x的平方/x) 30
4个回答
展开全部
(1+1/x)/[2x-(1+x^2)/x]
=[(x+1)/x]/{[2x^2-(1+x^2)]/x}
=[(x+1)/x]/{[2x^2-1-x^2)]/x}
=[(x+1)/x]/[(x^2-1)/x]
=(x+1)/x*x/(x^2-1)
=(x+1)/(x^2-1)
=(x+1)/(x-1)(x+1)
=1/(x-1) 希望能帮助你
=[(x+1)/x]/{[2x^2-(1+x^2)]/x}
=[(x+1)/x]/{[2x^2-1-x^2)]/x}
=[(x+1)/x]/[(x^2-1)/x]
=(x+1)/x*x/(x^2-1)
=(x+1)/(x^2-1)
=(x+1)/(x-1)(x+1)
=1/(x-1) 希望能帮助你
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
(1+1/x)/[2x-(1+x^2)/x]
=[(x+1)/x]/{[2x^2-(1+x^2)]/x}
=[(x+1)/x]/{[2x^2-1-x^2)]/x}
=[(x+1)/x]/[(x^2-1)/x]
=(x+1)/x*x/(x^2-1)
=(x+1)/(x^2-1)
=(x+1)/(x-1)(x+1)
=1/(x-1)
=[(x+1)/x]/{[2x^2-(1+x^2)]/x}
=[(x+1)/x]/{[2x^2-1-x^2)]/x}
=[(x+1)/x]/[(x^2-1)/x]
=(x+1)/x*x/(x^2-1)
=(x+1)/(x^2-1)
=(x+1)/(x-1)(x+1)
=1/(x-1)
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
2012-07-12
展开全部
1+1=2 会不会 就这样加上去 Sb
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