Question

解关于x的不等式（x平方-x+1）/（ax平方-ax-x+1）＞0...

Answer 1

由于x^2-x+1=(x-1/2)^2+3/4>0

故不等式就等价于ax^2-ax-x+1>0

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

(1)a>0:

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

(x-1)(x-1/a)>0

(i)a>1,1/a1或x