Answer
$\frac{5x-2}{(x+2)(x-2)(x-3)}$.
Work Step by Step
The given expression is
$=\frac{2}{x^2-5x+6}+\frac{3}{x^2-x-6}$
Factor $x^2-5x+6$.
Rewrite the middle term $-5x$ as $-3x-2x$
$=x^2-3x-2x+6$
Group terms.
$=(x^2-3x)+(-2x+6)$
Factor each term.
$=x(x-3)-2(x-3)$
Factor out $(x-3)$.
$=(x-3)(x-2)$.
Factor $x^2-x-6$.
Rewrite the middle term $-5x$ as $-3x+2x$
$=x^2-3x+2x-6$
Group terms.
$=(x^2-3x)+(2x-6)$
Factor each term.
$=x(x-3)+2(x-3)$
Factor out $(x-3)$.
$=(x-3)(x+2)$.
Substitute all factors into the given expression.
$=\frac{2}{(x-3)(x-2)}+\frac{3}{(x-3)(x+2)}$
LCM $=$ greatest power of all the prime factors.
LCM $=(x+2)(x-2)(x-3)$
$=\frac{(x+2)}{(x+2)}\cdot \frac{2}{(x-3)(x-2)}+\frac{(x-2)}{(x-2)}\cdot \frac{3}{(x-3)(x+2)}$
Simplify.
$=\frac{2(x+2)}{(x+2)(x-2)(x-3)}+ \frac{3(x-2)}{(x+2)(x-2)(x-3)}$
Add both numerators because both denominators are equal.
$=\frac{2(x+2)+3(x-2)}{(x+2)(x-2)(x-3)}$
Use the distributive property.
$=\frac{2x+4+3x-6}{(x+2)(x-2)(x-3)}$
Simplify.
$=\frac{5x-2}{(x+2)(x-2)(x-3)}$.
