解关于x的不等式a(x
原不等式可化为: [(a-1)x+(2-a)]/(x-2)>0 ∴(x-2)[(a-1)x+(2-a)]>0. (1) 当a>1时,有(x-2)[x-(a-2)/(a-1)]>0. 此时,(a-2)/(a-1)=1-1/(a-1)<2, 解集为{x|x>2,或x<(a-2)/(a-1)}. (2) 当a<1时,有(x-2)[x-(a-2)/(a-1)]<0. 若(a-2)/(a-1)>2,即0