c. \( \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} \)

To solve the equation \( \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} \), we first recognize that \( x^2 - 4 \) can be factored as \( (x - 2)(x + 2) \). This gives us the common denominator of \( (x - 2)(x + 2) \) for the left-hand side. Rewriting the equation with a common denominator, we get: \[ \frac{x+6 - 2(x+2)}{(x - 2)(x + 2)} = \frac{-1}{x + 2} \] Simplifying the numerator on the left, we combine terms: \[ x + 6 - 2x - 4 = -x + 2 \] This leads us to: \[ \frac{-x + 2}{(x - 2)(x + 2)} = \frac{-1}{x + 2} \] Now, cross-multiplying yields: \[ (-x + 2)(x + 2) = -1(x - 2)(x + 2) \] Expanding both sides gives: \[ -x^2 - 2x + 2x + 4 = -1(x^2 - 4) \] \[ -x^2 + 4 = -x^2 + 4 \] This confirms that both sides are equal, indicating that the equation holds for all \( x \) except where the denominators become zero. Therefore, \( x \) cannot be 2 or -2 due to the restrictions imposed by the original equation. So the solution is all real numbers except \( x = 2 \) and \( x = -2 \).