Solve $$ \frac{4 x}{x^{2}-9}=\frac{x-1}{x^{2}-6 x+9}+\frac{2}{x+3} $$ and give the extraneous solution. $$ x=\square $$ Extraneous solution $$ : x=\square $$

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$$
\textbf{Step 1: Factor the denominators.}
$$
The equation is
$$
\frac{4x}{x^2-9}=\frac{x-1}{x^2-6x+9}+\frac{2}{x+3}.
$$
Notice that
$$
x^2 - 9 = (x-3)(x+3) \quad \text{and} \quad x^2-6x+9 = (x-3)^2.
$$
Thus, the equation becomes
$$
\frac{4x}{(x-3)(x+3)} = \frac{x-1}{(x-3)^2} + \frac{2}{x+3}.
$$

$$
\textbf{Step 2: Identify restrictions.}
$$
The denominators $$ (x-3) $$, $$ (x+3) $$ must not be zero:
$$
x-3 \neq 0 \Rightarrow x \neq 3 \quad \text{and} \quad x+3 \neq 0 \Rightarrow x \neq -3.
$$

$$
\textbf{Step 3: Multiply through by the least common denominator (LCD).}
$$
The LCD of $$(x-3)(x+3)$$ and $$(x-3)^2$$ is
$$
(x-3)^2(x+3).
$$
Multiply each term by the LCD:
$$
4x \cdot \frac{(x-3)^2(x+3)}{(x-3)(x+3)} = (x-1) \cdot \frac{(x-3)^2(x+3)}{(x-3)^2} + 2 \cdot \frac{(x-3)^2(x+3)}{x+3}.
$$
Simplify each term:

- Left-hand side:
$$
4x \cdot \frac{(x-3)^2(x+3)}{(x-3)(x+3)} = 4x(x-3).
$$

- First term on the right:
$$
(x-1) \cdot \frac{(x-3)^2(x+3)}{(x-3)^2} = (x-1)(x+3).
$$

- Second term on the right:
$$
2 \cdot \frac{(x-3)^2(x+3)}{x+3} = 2(x-3)^2.
$$

Thus, the equation becomes
$$
4x(x-3) = (x-1)(x+3) + 2(x-3)^2.
$$

$$
\textbf{Step 4: Expand and simplify the equation.}
$$
Expand the left-hand side:
$$
4x(x-3) = 4x^2 - 12x.
$$
Expand the right-hand side:
$$
(x-1)(x+3) = x^2 + 2x - 3,
$$
$$
2(x-3)^2 = 2(x^2 - 6x + 9) = 2x^2 - 12x + 18.
$$
Adding the two expressions on the right:
$$
x^2 + 2x - 3 + 2x^2 - 12x + 18 = 3x^2 - 10x + 15.
$$
The equation becomes:
$$
4x^2 - 12x = 3x^2 - 10x + 15.
$$

$$
\textbf{Step 5: Rearranging the equation.}
$$
Subtract $$3x^2 - 10x + 15$$ from both sides:
$$
4x^2 - 12x - 3x^2 + 10x - 15 = 0.
$$
Combine like terms:
$$
x^2 - 2x - 15 = 0.
$$

$$
\textbf{Step 6: Solve the quadratic equation.}
$$
The quadratic equation is:
$$
x^2 - 2x - 15 = 0.
$$
We can factor this or use the quadratic formula. The discriminant is:
$$
\Delta = (-2)^2 - 4(1)(-15) = 4 + 60 = 64.
$$
Thus,
$$
x = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}.
$$
This gives the solutions:
$$
x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-6}{2} = -3.
$$

$$
\textbf{Step 7: Determine the extraneous solution.}
$$
Recall the restrictions: $$x \neq 3$$ and $$x \neq -3$$. The solution $$x = -3$$ must be rejected as extraneous.

$$
\boxed{x = 5, \quad \text{Extraneous solution: } x = -3.}
$$
$$ x = 5 $$, extraneous solution: $$ x = -3 $$.

Solve \( \frac{4 x}{x^{2}-9}=\frac{x-1}{x^{2}-6 x+9}+\frac{2}{x+3} \) and give the extraneous solution. \( x=\square \) Extraneous solution \( : x=\square \)

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