用换元法解方程(x-1)/(x+2)-3/(x-1)-1=0
展开全部
(x-1)/(x+2)-3/(x-1)-1=0
(x-1)/(x+2)-[(x+2)-(x-1)]/(x-1)-1=0
(x-1)/(x+2)-(x+2)/(x-1)+1-1=0
(x-1)/(x+2)-(x+2)/(x-1)=0
设(x-1)/(x+2)=y，则y-1/y=0
即：y²-1=0，解得：y1=1
y2=-1
即：(x-1)/(x+2)=1，或者(x-1)/(x+2)=-1
解得：x=-1/2
经检验是原方程的根
(x-1)/(x+2)-[(x+2)-(x-1)]/(x-1)-1=0
(x-1)/(x+2)-(x+2)/(x-1)+1-1=0
(x-1)/(x+2)-(x+2)/(x-1)=0
设(x-1)/(x+2)=y，则y-1/y=0
即：y²-1=0，解得：y1=1
y2=-1
即：(x-1)/(x+2)=1，或者(x-1)/(x+2)=-1
解得：x=-1/2
经检验是原方程的根
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
