若不等式x-a/x2+x+1＞x-b/x2-x+1的解为1/2＞x＞1,求a,b的值

x^2+x+1=(x+1/2)^2+3/4>0
x^2-x+1=(x-1/2)^2+3/4>0
两个分母都大于0
所以两边乘以(x^2+x+1)(x^2-x+1)
(x-a)(x^2-x+1)>(x-b)(x^2+x+1)
x^3-(a+1)x^2+(a+1)x-a>x^3+(1-b)x^2+(1-b)x-b
-(a+1)x^2+(a+1)x-a>(1-b)x^2+(1-b)x-b
(1-b+a+1)x+(1-b-a-1)x+a-b<0
(a-b+2)x^2-(a+b)x+(a-b)<0
解是1/2＞x＞1？？？
应该是1/2<x<1
则a-b+2>0
且(a-b+2)(x-1/2)(x-1)<0
(a-b+2)x^2-(3/2)(a-b+2)x+(1/2)(a-b+2)<0
对应项系数相等
所以-(3/2)(a-b+2)=-(a+b)
(1/2)(a-b+2)=a-b
即a-5b=-6 (1)
a-b=2 (2)
(2)-(1)
4b=8
b=2
a=2+b=4