Answer
$\dfrac{x-3}{2-x}\div\dfrac{x^{2}+3x-18}{x^{2}+2x-8}=-\dfrac{x+4}{x+6}$
Work Step by Step
$\dfrac{x-3}{2-x}\div\dfrac{x^{2}+3x-18}{x^{2}+2x-8}$
Factor the second fraction completely:
$\dfrac{x-3}{2-x}\div\dfrac{x^{2}+3x-18}{x^{2}+2x-8}=\dfrac{x-3}{2-x}\div\dfrac{(x+6)(x-3)}{(x+4)(x-2)}=...$
Evaluate the division of the two rational expressions:
$...=\dfrac{(x-3)(x+4)(x-2)}{(2-x)(x+6)(x-3)}=...$
Simplify by removing repeated factors in the numerator and the denominator. To remove $(x-2)$ and $(2-x)$, change the sign of the factor $(2-x)$ and also change the sign of the fraction:
$...=\dfrac{(x+4)(x-2)}{(2-x)(x+6)}=-\dfrac{(x+4)(x-2)}{(x-2)(x+6)}=-\dfrac{x+4}{x+6}$
