#### Answer

$\dfrac{x+2}{7-x}\div\dfrac{x^{2}-5x+6}{x^2-9x+14}=\dfrac{x+2}{3-x}$

#### Work Step by Step

$\dfrac{x+2}{7-x}\div\dfrac{x^{2}-5x+6}{x^2-9x+14}$
Factor the second fraction completely:
$\dfrac{x+2}{7-x}\div\dfrac{x^{2}-5x+6}{x^2-9x+14}=\dfrac{x+2}{7-x}\div\dfrac{(x-2)(x-3)}{(x-2)(x-7)}=...$
Evaluate the division of the two rational expressions:
$...=\dfrac{(x+2)(x-2)(x-7)}{(7-x)(x-2)(x-3)}=...$
Simplify by removing repeated factors in the numerator and the denominator. To remove $(x-7)$ and $(7-x)$, change the sign of the factor $(7-x)$ and change the sign of the fraction:
$...=\dfrac{(x+2)(x-7)}{(7-x)(x-3)}=-\dfrac{(x-7)(x+2)}{(x-3)(x-7)}=-\dfrac{x+2}{x-3}=\dfrac{x+2}{3-x}$