Simplify (1/9-1/(x^2))/(3+9/x)

$\frac{\frac{1}{9} - \frac{1}{x^{2}}}{3 + \frac{9}{x}}$

Multiply the numerator and denominator of the complex fraction by $9 x^{2}$ .

Multiply $\frac{\frac{1}{9} - \frac{1}{x^{2}}}{3 + \frac{9}{x}}$ by $\frac{9 x^{2}}{9 x^{2}}$ .

$\frac{9 x^{2}}{9 x^{2}} \cdot \frac{\frac{1}{9} - \frac{1}{x^{2}}}{3 + \frac{9}{x}}$

Combine.

$\frac{9 x^{2} (\frac{1}{9} - \frac{1}{x^{2}})}{9 x^{2} (3 + \frac{9}{x})}$

Apply the distributive property.

$\frac{9 x^{2} \frac{1}{9} + 9 x^{2} (- \frac{1}{x^{2}})}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Simplify by cancelling.

Cancel the common factor of $9$ .

Factor $9$ out of $9 x^{2}$ .

$\frac{9 (x^{2}) \frac{1}{9} + 9 x^{2} (- \frac{1}{x^{2}})}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Cancel the common factor.

$\frac{9 x^{2} \frac{1}{9} + 9 x^{2} (- \frac{1}{x^{2}})}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Rewrite the expression.

$\frac{x^{2} + 9 x^{2} (- \frac{1}{x^{2}})}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Cancel the common factor of $x^{2}$ .

Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

$\frac{x^{2} + 9 x^{2} \frac{- 1}{x^{2}}}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Factor $x^{2}$ out of $9 x^{2}$ .

$\frac{x^{2} + x^{2} \cdot 9 \frac{- 1}{x^{2}}}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Cancel the common factor.

$\frac{x^{2} + x^{2} \cdot 9 \frac{- 1}{x^{2}}}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Rewrite the expression.

$\frac{x^{2} + 9 \cdot - 1}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Multiply $9$ by $- 1$ .

$\frac{x^{2} - 9}{9 x^{2} \cdot 3 + 9 x^{2} \frac{9}{x}}$

Cancel the common factor of $x$ .

Factor $x$ out of $9 x^{2}$ .

$\frac{x^{2} - 9}{9 x^{2} \cdot 3 + x (9 x) \frac{9}{x}}$

Cancel the common factor.

$\frac{x^{2} - 9}{9 x^{2} \cdot 3 + x (9 x) \frac{9}{x}}$

Rewrite the expression.

$\frac{x^{2} - 9}{9 x^{2} \cdot 3 + 9 x \cdot 9}$

Multiply $9$ by $9$ .

$\frac{x^{2} - 9}{9 x^{2} \cdot 3 + 81 x}$

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$\frac{x^{2} - 3^{2}}{9 x^{2} \cdot 3 + 81 x}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{(x + 3) (x - 3)}{9 x^{2} \cdot 3 + 81 x}$

Simplify the denominator.

Factor $9 x \cdot 3$ out of $9 x^{2} \cdot 3 + 81 x$ .

Factor $9 x \cdot 3$ out of $9 x^{2} \cdot 3$ .

$\frac{(x + 3) (x - 3)}{9 x \cdot 3 (x) + 81 x}$

Factor $9 x \cdot 3$ out of $81 x$ .

$\frac{(x + 3) (x - 3)}{9 x \cdot 3 x + 9 x \cdot 3 \cdot 3}$

Factor $9 x \cdot 3$ out of $9 x \cdot 3 x + 9 x \cdot 3 \cdot 3$ .

$\frac{(x + 3) (x - 3)}{9 x \cdot 3 (x + 3)}$

Multiply $3$ by $9$ .

$\frac{(x + 3) (x - 3)}{27 x (x + 3)}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{(x + 3) (x - 3)}{27 x (x + 3)}$

Rewrite the expression.

$\frac{x - 3}{27 x}$

Simplify (1/9-1/(x^2))/(3+9/x)

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