(2x/x²-4-1/x-2)×x+2/x-1
5个回答
展开全部
原式=[2x/(x+2)(x-2)-1/(x-2)]×(x+2)/(x+1)
=(2x-x-2)/(x+2)(x-2)×(x+2)/(x+1)
=(x-2)/(x+2)(x-2)×(x+2)/(x+1)
=1/(x+1)
=(2x-x-2)/(x+2)(x-2)×(x+2)/(x+1)
=(x-2)/(x+2)(x-2)×(x+2)/(x+1)
=1/(x+1)
追问
答案是1/x-1,请再看一下
追答
哦,对不起,我写错了
吧x-1看成x+1了
原式=[2x/(x+2)(x-2)-1/(x-2)]×(x+2)/(x-1)
=(2x-x-2)/(x+2)(x-2)×(x+2)/(x-1)
=(x-2)/(x+2)(x-2)×(x+2)/(x-1)
=1/(x-1)
展开全部
{[(x+2)+(x-2)]/[(x+2)(x-2)]-1/(x-2)}(x+2)/(x-1)=[1/(x-2)+1/(x+2)-1/(x-2)](x+2)/(x-1)=[1/(x+2)](x+2)/(x-1)=1/(x-1). 分不清楚括号在哪儿,只能猜测了,呵呵。
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
(2x/x²-4-1/x-2)×x+2/x-1
=[2x/(x+2)(x-2)-1/(x-2)]×(x+2)/(x-1)
=[2x/(x+2)(x-2)-(x+2)/(x+2)(x-2)]×(x+2)/(x-1)
=[(2x-(x-2))/(x+2)(x-2)]×(x+2)/(x-1)
=(x-2)/(x+2)(x-2)×(x+2)/(x-1)
=1/(x+2)×(x+2)/(x-1)
=1/(x-1)
=[2x/(x+2)(x-2)-1/(x-2)]×(x+2)/(x-1)
=[2x/(x+2)(x-2)-(x+2)/(x+2)(x-2)]×(x+2)/(x-1)
=[(2x-(x-2))/(x+2)(x-2)]×(x+2)/(x-1)
=(x-2)/(x+2)(x-2)×(x+2)/(x-1)
=1/(x+2)×(x+2)/(x-1)
=1/(x-1)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
10
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
广告 您可能关注的内容 |