Answer
The simplified form of a complex rational expression $\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}$ is $x-3$.
Work Step by Step
$\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}$
The denominators within the complex rational expression are $x\text{ and }3$.
Thus, the LCD is $3x$.
Multiply the expression with $\frac{3x}{3x}$:
$\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}\cdot \frac{3x}{3x}$
Apply the distributive property,
$\begin{align}
& \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\frac{x}{3}\cdot 3x-\frac{3}{x}\cdot 3x}{\frac{1}{x}\cdot 3x+\frac{1}{3}\cdot 3x} \\
& =\frac{x\cdot x-3\cdot 3}{1\cdot 3+1\cdot x}
\end{align}$
Simplify the terms,
$\begin{align}
& \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{{{x}^{2}}-9}{3+x} \\
& =\frac{{{x}^{2}}-9}{x+3}
\end{align}$
$\begin{align}
& \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\left( x+3 \right)\left( x-3 \right)}{x+3} \\
& =x-3
\end{align}$
