2х^2+х-1/х^2-1=0

To solve the equation:

\[ \frac{2x^2 + x - 1}{x^2 - 1} = 0 \]

we set the numerator equal to zero, since a fraction is zero when its numerator is zero (provided the denominator is not zero).

1. Set the numerator equal to zero:

\[ 2x^2 + x - 1 = 0 \]

2. Now, we can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \(a = 2\), \(b = 1\), and \(c = -1\).

Calculate the discriminant:

\[ b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 \]

Now use the quadratic formula:

\[ x = \frac{-1 \pm \sqrt{9}}{2 \cdot 2} = \frac{-1 \pm 3}{4} \]

3. This gives us two solutions:

\[ x_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \] \[ x_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \]

4. However, we must also ensure that these values don't make the denominator zero:

The denominator is \(x^2 - 1\). Set it to zero to check for restrictions:

\[ x^2 - 1 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \]

So \(x = -1\) is not allowed because it makes the denominator zero.

Thus, the only valid solution is:

\[ \boxed{\frac{1}{2}} \]