$$ \frac{x_{9}^{2}}{9}+\frac{36}{x^{2}} $$, if $$ \frac{x}{3}-\frac{6}{x}=1 $$

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Given the equation $$ \frac{x}{3}-\frac{6}{x}=1 $$, we need to solve for $$ x $$ first.

Let's solve the equation $$ \frac{x}{3}-\frac{6}{x}=1 $$ for $$ x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{x}{3}-\frac{6}{x}=1$$
- step1: Find the domain:
  $$\frac{x}{3}-\frac{6}{x}=1,x
eq 0$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{x}{3}-\frac{6}{x}ight)	imes 3x=1	imes 3x$$
- step3: Simplify the equation:
  $$x^{2}-18=3x$$
- step4: Move the expression to the left side:
  $$x^{2}-18-3x=0$$
- step5: Factor the expression:
  $$\left(x-6ight)\left(x+3ight)=0$$
- step6: Separate into possible cases:
  $$\begin{align}&x-6=0\&x+3=0\end{align}$$
- step7: Solve the equation:
  $$\begin{align}&x=6\&x=-3\end{align}$$
- step8: Check if the solution is in the defined range:
  $$\begin{align}&x=6\&x=-3\end{align},x
eq 0$$
- step9: Find the intersection:
  $$\begin{align}&x=6\&x=-3\end{align}$$
- step10: Rewrite:
  $$x_{1}=-3,x_{2}=6$$
The solutions to the equation $$ \frac{x}{3}-\frac{6}{x}=1 $$ are $$ x = -3 $$ and $$ x = 6 $$.

Now, we need to substitute these values of $$ x $$ into the expression $$ \frac{x_{9}^{2}}{9}+\frac{36}{x^{2}} $$ to find the final result.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{x_{9}^{2}}{9}+\frac{36}{x^{2}}$$
- step1: Reduce fractions to a common denominator:
  $$\frac{x_{9}^{2}x^{2}}{9x^{2}}+\frac{36	imes 9}{x^{2}	imes 9}$$
- step2: Reorder the terms:
  $$\frac{x_{9}^{2}x^{2}}{9x^{2}}+\frac{36	imes 9}{9x^{2}}$$
- step3: Transform the expression:
  $$\frac{x_{9}^{2}x^{2}+36	imes 9}{9x^{2}}$$
- step4: Multiply the numbers:
  $$\frac{x_{9}^{2}x^{2}+324}{9x^{2}}$$
Substitute $$ x = -3 $$ into the expression $$ \frac{x_{9}^{2}}{9}+\frac{36}{x^{2}} $$:

$$ \frac{(-3)^{2}}{9}+\frac{36}{(-3)^{2}} = \frac{9}{9}+\frac{36}{9} = 1+4 = 5 $$

Substitute $$ x = 6 $$ into the expression $$ \frac{x_{9}^{2}}{9}+\frac{36}{x^{2}} $$:

$$ \frac{6^{2}}{9}+\frac{36}{6^{2}} = \frac{36}{9}+\frac{36}{36} = 4+1 = 5 $$

Therefore, the final result for both $$ x = -3 $$ and $$ x = 6 $$ is 5.
When $$ x = -3 $$ or $$ x = 6 $$, the expression $$ \frac{x_{9}^{2}}{9}+\frac{36}{x^{2}} $$ equals 5.

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