Solve \( \begin{array}{l}\frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{x^{2}-4} \\ x=-1 \\ x=-2 \\ x=4 \\ x=2 \& 4\end{array} \)

To solve the equation \( \frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{x^{2}-4} \), first, recognize that \( x^2 - 4 = (x+2)(x-2) \). Thus, the right side can be rewritten as: \[ \frac{x-1}{(x+2)(x-2)}. \] Now, rewrite the left side with a common denominator of \( (x+2)(x-2) \): \[ \frac{2(x-2) + 4(x+2)}{(x+2)(x-2)} = \frac{2x - 4 + 4x + 8}{(x+2)(x-2)} = \frac{6x + 4}{(x+2)(x-2)}. \] Now we have: \[ \frac{6x + 4}{(x+2)(x-2)} = \frac{x-1}{(x+2)(x-2)}. \] Since the denominators are equal, we can set the numerators equal to each other: \[ 6x + 4 = x - 1. \] Now, solve for \( x \): \[ 6x - x = -1 - 4 \\ 5x = -5 \\ x = -1. \] Now we need to check if any values make the original fractions undefined. The denominators \( x + 2 \) and \( x - 2 \) would be undefined for: - \( x = -2 \) (undefined) - \( x = 2 \) (undefined) Thus, the only solution to the equation is: \[ x = -1. \] Checking the other provided options: \( x = -2 \) and \( x = 2 \) are not valid, and \( x = 4 \) does not satisfy the equation either. So, the final solution is: \[ \boxed{-1}. \]