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. Use the quadratic formula to solve for \(x\): The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 2\), \(b = 1\), and \(c = -1\).

3. Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9 \]

4. Find the roots: \[ x = \frac{-1 \pm \sqrt{9}}{2 \times 2} = \frac{-1 \pm 3}{4} \]

   This gives us two possible solutions: \[ x_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \] \[ x_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \]

5. Check the denominator to ensure these values do not make it zero: The denominator is: \[ x^2 - 1 = 0 \implies x = \pm 1 \]

   Since \(x = -1\) would make the denominator zero, it is not a valid solution.

Thus, the only solution to the equation \(\frac{2x^2 + x - 1}{x^2 - 1} = 0\) is: \[ \boxed{\frac{1}{2}} \]