To solve the equation
\[ \frac{7}{3-x} = \frac{5}{x+3} + \frac{4}{x^2 - 9}, \]
we can start by noting that the term \(x^2 - 9\) can be factored as \((x - 3)(x + 3)\). Thus, we rewrite the equation:
\[ \frac{7}{3-x} = \frac{5}{x+3} + \frac{4}{(x-3)(x+3)}. \]
Cross-multiplying to eliminate the fractions is often a good strategy. Let's first find a common denominator for the right side of the equation.
The common denominator for the right-hand side is \((x - 3)(x + 3)\). Therefore, we can rewrite the right side:
\[ \frac{5}{x+3} = \frac{5(x-3)}{(x-3)(x+3)}, \]
yielding:
\[ \frac{5(x-3) + 4}{(x-3)(x+3)} = \frac{5x - 15 + 4}{(x-3)(x+3)} = \frac{5x - 11}{(x-3)(x+3)}. \]
Now our equation looks like this:
\[ \frac{7}{3-x} = \frac{5x - 11}{(x-3)(x+3)}. \]
To make it easier to work with, we can rewrite \(3 - x\) as \(-(x - 3)\):
\[ \frac{7}{-(x - 3)} = \frac{5x - 11}{(x-3)(x + 3)}. \]
This can be rearranged to:
\[ -\frac{7}{x - 3} = \frac{5x - 11}{(x - 3)(x + 3)}. \]
Now, we can multiply both sides by \((x - 3)(x + 3)\) to eliminate the denominators (assuming \(x \neq 3\) and \(x \neq -3\)):
\[ -7(x + 3) = 5x - 11. \]
Expanding the left side gives:
\[ -7x - 21 = 5x - 11. \]
Next, we want to isolate \(x\). First, add \(7x\) to both sides:
\[ -21 = 12x - 11. \]
Now add \(11\) to both sides:
\[ -10 = 12x. \]
Dividing by \(12\):
\[ x = -\frac{10}{12} = -\frac{5}{6}. \]
We should check if this solution satisfies the original equation and that it does not cause any zeros in the denominators. For \(x = -\frac{5}{6}\):
- \(3-x = 3 + \frac{5}{6} = \frac{18}{6} + \frac{5}{6} = \frac{23}{6}\) (not zero),
- \(x+3 = -\frac{5}{6} + 3 = -\frac{5}{6} + \frac{18}{6} = \frac{13}{6}\) (not zero),
- \(x^2 - 9 = \left(-\frac{5}{6}\right)^2 - 9 = \frac{25}{36} - 9 = \frac{25}{36} - \frac{324}{36} = -\frac{299}{36}\) (not zero).
Thus, the solution is valid. Therefore, the final answer is:
\[ \boxed{-\frac{5}{6}}. \]