二次函数、方程、不等式的关系2

a(x-1)/(x-2)>1
(ax-a)/(x-2)-1>0
(ax-a-x+2)/(x-2)>0
所以(ax-a-x+2)(x-2)>0
[(a-1)x-(a-2)](x-2)>0

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

若a-1=0,a=1
则不等式是1*(x-2)>0
x>2

若a-1<0,a<1
两边除a-1
[x-(a-2)/(a-1)](x-2)<0
(a-2)/(a-1)-2=(a-2-2a+2)/(a-1)=-a/(a-1)
因为a-1<0,则0<a<1时，-a<0,
(a-2)/(a-1)-2<0
(a-2)/(a-1)<2
所以(a-2)/(a-1)<x<2
若a=0,则a(x-1)/x-2＞1是0>1,不成立
若a<0,-a/(a-1)>0
(a-2)/(a-1)-2>0
(a-2)/(a-1)>2
所以2<x<(a-2)/(a-1)

综上
a>1,x<(a-2)/(a-1),x>2
a=1,x>2
0<a<1,(a-2)/(a-1)<x<2
a=0,无解
a<0,2<x<(a-2)/(a-1)

a(x-1)/x-2＞1可化为a(x-1)/x-3>0的[(a-3)x-3]/x>0.
两根为3/