数学题目,分式的化简:1/a-1 + 1/(a-1)(a-2) + 1/(a-2)(a-3) +.....+ 1/(a-99)(a-100)
1个回答
展开全部
1=(a-1)-(a-2)
所以1/(a-1)(a-2)=[(a-1)-(a-2)]/(a-1)(a-2)=(a-1)/(a-1)(a-2)-(a-2)/(a-1)(a-2)=1/(a-2)-1/(a-1)
同理,1=(a-2)-(a-3),得到1/(a-2)(a-3)=1/(a-3)/1/(a-2)
后面以此类推
所以原式=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)/1/(a-2)+……+1/(a-100)-1/(a-99)
=1/(a-100)
所以1/(a-1)(a-2)=[(a-1)-(a-2)]/(a-1)(a-2)=(a-1)/(a-1)(a-2)-(a-2)/(a-1)(a-2)=1/(a-2)-1/(a-1)
同理,1=(a-2)-(a-3),得到1/(a-2)(a-3)=1/(a-3)/1/(a-2)
后面以此类推
所以原式=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)/1/(a-2)+……+1/(a-100)-1/(a-99)
=1/(a-100)
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