Mathematics
OpenStudy (anonymous):

$\Large \frac{1}{x^2+5x+6}+\frac{1}{x+3}$ $\Large \frac{1}{(x+3)(x+2)}+\frac{1}{x+3}$ $\Large \frac{1}{(x+3)(x+2)}+\frac{1(x+2)}{(x+3)(x+2)}$ $\Large \frac{1}{(x+3)(x+2)}+\frac{x+2}{(x+3)(x+2)}$ $\Large \frac{1+x+2}{(x+3)(x+2)}$ $\Large \frac{x+3}{(x+3)(x+2)}$ $\Large \frac{1}{x+2}$ So $\Large \frac{1}{x^2+5x+6}+\frac{1}{x+3}=\frac{1}{x+2}$ for all values of x where x can't equal -2 or can't equal -3