Answer
$ 6$.
Work Step by Step
The given expression is
$\Rightarrow \frac{3}{1-\frac{3}{3+x}}-\frac{3}{\frac{3}{3-x}-1}$
Solve the denominator of the first fraction.
$=1-\frac{3}{3+x}$
$=\frac{1}{1}-\frac{3}{3+x}$
The LCD of the denominators is $(3+x)$.
$=\frac{3+x}{3+x}-\frac{3}{3+x}$
$=\frac{3+x-3}{3+x}$
Simplify.
$=\frac{x}{3+x}$
Solve the denominator of the second fraction.
$=\frac{3}{3-x}-1$
$=\frac{3}{3-x}-\frac{1}{1}$
The LCD of the denominators is $(3-x)$.
$=\frac{3}{3-x}-\frac{3-x}{3-x}$
$=\frac{3-(3-x)}{3-x}$
Simplify.
$=\frac{3-3+x}{3-x}$
$=\frac{x}{3-x}$
Back substitute all values into the given fraction.
$\Rightarrow \frac{3}{\frac{x}{3+x}}-\frac{3}{\frac{x}{3-x}}$
Invert the divisor and multiply.
$\Rightarrow 3\cdot \frac{3+x}{x}-3\cdot \frac{3-x}{x}$
$\Rightarrow \frac{3(3+x)}{x}-\frac{3(3-x)}{x}$
$\Rightarrow \frac{9+3x}{x}-\frac{9-3x}{x}$
$\Rightarrow \frac{9+3x-(9-3x)}{x}$
$\Rightarrow \frac{9+3x-9+3x}{x}$
Simplify.
$\Rightarrow \frac{6x}{x}$
Cancel common terms.
$\Rightarrow 6$.
