2х^2+х-1/х^2-1=0

To solve the equation \(\frac{2x^2 + x - 1}{x^2 - 1} = 0\), we can follow these steps:

1. Set the numerator equal to zero: Since a fraction is equal to zero when its numerator is zero (provided the denominator is not zero), we set the numerator \(2x^2 + x - 1\) equal to zero:

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

2. Use the quadratic formula: The quadratic formula is given by:

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

   Here, \(a = 2\), \(b = 1\), and \(c = -1\). We first calculate the discriminant:

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

   Now substitute the values into the quadratic formula:

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

   This gives us two possible solutions:

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

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

3. Check for restrictions from the denominator: The denominator \(x^2 - 1\) cannot be zero. Therefore, we need to check if any of our solutions lead to a zero denominator:

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

   Since \(x = -1\) is a solution found in step 2, it must be discarded as it makes the denominator zero.

4. Final Solution: Thus, the only valid solution is:

   \[ x = \frac{1}{2} \]

Therefore, the solution to the equation \(\frac{2x^2 + x - 1}{x^2 - 1} = 0\) is \(x = \frac{1}{2}\).