早教吧作业答案频道 -->数学-->
用换元法解方程(x-1)/(x+2)-3/(x-1)-1=0
题目详情
用换元法解方程(x-1)/(x+2)-3/(x-1)-1=0
▼优质解答
答案和解析
(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
看了 用换元法解方程(x-1)/(...的网友还看了以下: