Not directly. You may try :

```
sage: S1=solve_ineq([(a-b*x^2)/(x-1)>0],[x]);S1
[[x < 1, b*x^2 - a > 0], [1 < x, -b*x^2 + a > 0]]
```

But there, you're on your own:

```
sage: solve_ineq([(S1[0][1]+a)/b],[x])
[[b > 0, b*x^2 - a > 0], [-b > 0, -b*x^2 + a > 0]]
```

Sage may help you playing with that :

```
sage: S2=(S1[0][1]+a)/b; S2
x^2 > a/b
```

but

```
sage: solve_ineq([S2],[x])
[[b > 0, b*x^2 - a > 0], [-b > 0, -b*x^2 + a > 0]]
```

is pretty unhelpful.

Note that "the competition" isn't very helpful, either :

```
sage: mathematica("Reduce[Element[a, Reals] && Element[b, Reals] && Element[x, Reals] && (a-b*x^2)/(x-1)>0 ,x]")
(b < 0 && ((a < b && (Inequality[-Sqrt[a/b], Less, x, Less, 1] ||
x > Sqrt[a/b])) || (a == b && (Inequality[-Sqrt[a/b], Less, x, Less,
1] || x > 1)) || (Inequality[b, Less, a, LessEqual, 0] &&
(Inequality[-Sqrt[a/b], Less, x, Less, Sqrt[a/b]] || x > 1)) ||
(a > 0 && x > 1))) || (b == 0 && ((a < 0 && x < 1) ||
(a > 0 && x > 1))) || (b > 0 && ((a < 0 && x < 1) ||
(Inequality[0, LessEqual, a, Less, b] && (x < -Sqrt[a/b] ||
Inequality[Sqrt[a/b], Less, x, Less, 1])) ||
(a == b && x < -Sqrt[a/b]) || (a > b && (x < -Sqrt[a/b] ||
Inequality[1, Less, x, Less, Sqrt[a/b]]))))
```

Current progress in `sympy`

may someday get us more helpful answere, but we're not here yet.

Another possibility is to use one of the linear prograpping packages available in Sage, bu I do not know them well enough to point you in the right direction...

HTH,