2) \( \frac{x^{2}-9}{x^{2}-5 x+6} \div \frac{3-x}{x+2} \)

To solve the expression \( \frac{x^{2}-9}{x^{2}-5 x+6} \div \frac{3-x}{x+2} \), let's start by rewriting the division as multiplication by the reciprocal. This gives us: \[ \frac{x^{2} - 9}{x^{2} - 5x + 6} \times \frac{x + 2}{3 - x} \] Next, let's factor everything we can. The numerator \( x^2 - 9 \) is a difference of squares: \[ x^{2} - 9 = (x - 3)(x + 3) \] The denominator \( x^{2} - 5x + 6 \) can be factored into: \[ x^{2} - 5x + 6 = (x - 2)(x - 3) \] Now, we also notice that \( 3 - x \) can be rewritten as \( -(x - 3) \). With these factors, our expression now looks like: \[ \frac{(x - 3)(x + 3)}{(x - 2)(x - 3)} \times \frac{x + 2}{-(x - 3)} \] Now we can cancel the \( (x - 3) \) terms: \[ \frac{(x + 3)(x + 2)}{-(x - 2)} \] This simplifies to: \[ -\frac{(x + 3)(x + 2)}{x - 2} \] So the final answer is: \[ -\frac{(x + 3)(x + 2)}{x - 2} \]