Question

解关于x的不等式[(a-1)x+2-a](x-2)>0.

需要详细解答过程。

Answer 1

若a-1=0
a=1
则不等式为1*(x-2)>0
x>2

若a-1>0，a>1
则两边除a-1
[x+(2-a)/(a-1)](x-2)>0
[x-(a-2)/(a-1)](x-2)>0
(a-2)/(a-1)=(a-1-1)/(a-1)=1-1/(a-1)
a-1>0,1/(a-1)>0,-1/(a-1)<0
所以1-1/(a-1)<1<2
所以x>2,x<(a-2)/(a-1)

若a-1<0，a<1
则两边除a-1
[x+(2-a)/(a-1)](x-2)<0
[x-(a-2)/(a-1)](x-2)<0
(a-2)/(a-1)=(a-1-1)/(a-1)=1-1/(a-1)
若0<a<1,则-1<a-1<0
1/(a-1)<-1,-1/(a-1)>1
1-1/(a-1)>2
则2<x<(a-2)/(a-1)
若a=0,则(a-2)/(a-1)=2，
不等式为(x-2)^2<0,不成立，无解
若a<0,则a-1<-1
-1<1/(a-1)<0,0<-1/(a-1)<1
1<1-1/(a-1)<2
则(a-2)/(a-1)<x<2