(a-1)x^2-(a-2)x-1>0

当a=1时 ,x>1
当a!=1时 由十字相乘法可知
令原式等式为0
所以
[(a-1)x+1]x(x-1)=0
x1=1/1-a,x2=1;
综上所述可知
当01/1-a或x当a1时解为1/1-a

1、 当a=1时,原不等式变为 x-1>0,即 x > 1
2、当a不等于1时,x^2 - [(a-1)/(a-1)]x - 1/(a-1) >0
x^2 - 2 [(a-2)/2(a-1)]x + [(a-1)/2(a-1)]^2 -[(a-1)/2(a-1)]^2 - 1/(a-1) >0
[ x- (a-2)/2(a-1)]^2 - [a/2(a-1)]^2 >0
[ x- (a-2)/2(a-1)]^2 > [a/2(a-1)]^2
分两种情况讨论
(1) x- (a-2)/2(a-1) > a/2(a-1)
x > a/2(a-1) + (a-2)/2(a-1)
x > 1
(2) x- (a-2)/2(a-1) >-[ a/2(a-1)]
x > (a-2)/2(a-1) - [ a/2(a-1)]
x > -1/(a-1) = 1/(1-a)
综上,x >1 或x >1/(1-a)
此类不等式的解答思路：先分情况讨论二次项系数是否为0,然后再详细讨论二次项不为0的情况