Answer
$ \frac{x-3}{x}$.
Work Step by Step
The given expression has fractions both in the numerator and denominator. To simplify, we multiply both numerator and denominator of the given expression by the Least Common denominator of the fractions, which is $LCD=x^2$:
$\frac{1-\frac{9}{x^2}}{1+\frac{3}{x}}=\frac{x^2}{x^2}\cdot \frac{1-\frac{9}{x^2}}{1+\frac{3}{x}}$
Use the distributive property.
$= \frac{x^2\cdot 1-x^2\cdot \frac{9}{x^2}}{x^2\cdot1+x^2\cdot\frac{3}{x}}$
Simplify.
$= \frac{x^2-9}{x^2+3x}$
Factor the numerator $x^2-9$.
$=x^2-3^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+3)(x-3)$
Factor the denominator $x^2+3x$.
$=x(x+3)$.
Substitute the factors into the fraction.
$ =\frac{(x+3)(x-3)}{x(x+3)}$
Cancel the common term.
$ =\frac{x-3}{x}$.
