Answer
$\frac{2x^2-9}{(x+3)(x-3)}$.
Work Step by Step
The given expression is
$=\frac{x}{x+3}+\frac{x}{x-3}-\frac{9}{x^2-9}$
Factor $x^2-9$
$=x^2-3^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+3)(x-3)$.
Substitute the factor into the given expression.
$=\frac{x}{x+3}+\frac{x}{x-3}-\frac{9}{(x+3)(x-3)}$
LCM $=(x+3)(x-3)$
$=\frac{(x-3)}{(x-3)}\cdot \frac{x}{x+3}+\frac{(x+3)}{(x+3)}\cdot \frac{x}{x-3}-\frac{9}{(x+3)(x-3)}$
Simplify.
$=\frac{x(x-3)}{(x+3)(x-3)}+ \frac{x(x+3)}{(x-3)(x+3)}-\frac{9}{(x+3)(x-3)}$
Add both numerators because both denominators are equal.
$=\frac{x(x-3)+x(x+3)-9}{(x+3)(x-3)}$
Use the distributive property in the numerator.
$=\frac{x^2-3x+x^2+3x-9}{(x+3)(x-3)}$
Simplify.
$=\frac{2x^2-9}{(x+3)(x-3)}$.
