Answer
$-\frac{x-4}{x+3}$ and $x\ne3$.
Work Step by Step
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $\frac{x^2-7x+12}{9-x^2}=\frac{x^2-3x-4x-12}{-(x^2-3^2)}=\frac{x(x-3)-4(x-3)}{-(x+3)(x-3)}=\frac{(x-3)(x-4)}{-(x+3)(x-3)}=-\frac{x-4}{x+3}$ and $x\ne3$.
