Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 41

$\frac{1}{x+3}$

$x^2+9x+18=x^2+3x+6x+18=x(x+3)+6(x+3)=(x+3)(x+6)$, so LCD$=(x+3)(x+6)$, thus $\frac{x^2-6}{x^2+9x+18}-\frac{x-4}{x+6}=\frac{x^2-6}{(x+3)(x+6)}-\frac{(x-4)(x+3)}{(x+3)(x+6)}=\frac{x^2-6-(x^2-x-12)}{(x+3)(x+6)}=\frac{6+x}{(x+3)(x+6)}=\frac{1}{x+3}$