1、 當(dāng)a=1時(shí),原不等式變?yōu)?x-1>0,即 x > 1
2、當(dāng)a不等于1時(shí),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
分兩種情況討論
(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)
綜上,x >1 或x >1/(1-a)
此類(lèi)不等式的解答思路：先分情況討論二次項(xiàng)系數(shù)是否為0,然后再詳細(xì)討論二次項(xiàng)不為0的情況