Solve the following inequalities:
`(x^(2)-3x-18)/(13x-x^(2)-42)gt0`

Answer

To solve the inequality \((x^2 - 3x - 18)/(13x - x^2 - 42) > 0\), we will follow these steps:

### Step 1: Factor the numerator and denominator

1. **Numerator**:
We need to factor \(x^2 - 3x - 18\). This can be factored as:
\[ x^2 - 3x - 18 = (x - 6)(x + 3) \]

2. **Denominator**:
We rewrite \(13x - x^2 - 42\) as \(-x^2 + 13x - 42\) or \( -(x^2 - 13x + 42)\). Now we factor \(x^2 - 13x + 42\):
\[ x^2 - 13x + 42 = (x - 6)(x - 7) \]
Thus, the denominator becomes:
\[ -(x - 6)(x - 7) \]

### Step 2: Rewrite the inequality

Now substituting the factored forms into the inequality, we have:
\[ \frac{(x - 6)(x + 3)}{-(x - 6)(x - 7)} > 0 \]

### Step 3: Simplify the inequality

We can simplify this expression:
\[ \frac{(x - 6)(x + 3)}{-(x - 6)(x - 7)} = \frac{(x + 3)}{-(x - 7)} \]
However, we must note that \(x \neq 6\) because it would make the denominator zero.

### Step 4: Analyze the sign of the expression

Next, we need to determine where the expression \(\frac{(x + 3)}{-(x - 7)} > 0\) holds true.

1. The critical points are \(x = -3\) and \(x = 7\).
2. We will test the intervals determined by these points:
- \( (-\infty, -3) \)
- \( (-3, 6) \)
- \( (6, 7) \)
- \( (7, \infty) \)

### Step 5: Test the intervals

1. **For \(x < -3\)** (e.g., \(x = -4\)):
\[ \frac{(-4 + 3)}{-( -4 - 7)} = \frac{-1}{11} < 0 \]

2. **For \(-3 < x < 6\)** (e.g., \(x = 0\)):
\[ \frac{(0 + 3)}{-(0 - 7)} = \frac{3}{7} > 0 \]

3. **For \(6 < x < 7\)** (e.g., \(x = 6.5\)):
\[ \frac{(6.5 + 3)}{-(6.5 - 7)} = \frac{9.5}{-0.5} < 0 \]

4. **For \(x > 7\)** (e.g., \(x = 8\)):
\[ \frac{(8 + 3)}{-(8 - 7)} = \frac{11}{-1} < 0 \]

### Step 6: Combine the results

From our tests, the expression is positive in the interval:
\[ (-3, 6) \]
And we must exclude \(x = 6\) since it makes the denominator zero.

### Final Solution:

Thus, the solution to the inequality is:
\[ \boxed{(-3, 6)} \]