1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy

[复制链接]
查看11 | 回复1 | 2012-9-2 17:36:13 | 显示全部楼层 |阅读模式
解1题:原式=[1/(a+1)]-[2/(1-a)]=[1/(a+1)]+[2/(a-1)]={ (a-1)/[(a+1)(a-1)] }+{2(a+1)/[(a+1)(a-1)] }=[(a-1)+2(a+1)]/[(a+1)(a-1)]=(a-1+2a+2)/(a2-1)=(3a+1)/(a2-1) 解2题:原式=[x/(x-y)]-[y/(x+y)]={ x(x+y)/[(x+y)(x-y)] }-{ y(x-y)/[(x+y)(x-y)] }=[x(x+y)-y(x-y)]/[(x+y)(x-y)]=(x2+xy-xy+y2)/(x2-y&#178...
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 立即注册

本版积分规则

主题

0

回帖

4882万

积分

论坛元老

Rank: 8Rank: 8

积分
48824836
热门排行