### Instruction: Solve
\[\frac{|x - 13|}{x^2 - 9x - 52} \le 0.\].### Response: To solve an inequality involving absolute value, I need to consider two cases: when the expression inside the absolute value is positive, and when it is negative.
For the first case, I can assume that $x - 13 \ge 0$, or equivalently, $x \ge 13$.
Then, the inequality becomes
\[\frac{x - 13}{x^2 - 9x - 52} \le 0.\]
To simplify this, I can factor the denominator as $(x - 13)(x + 4)$, and cancel out the common factor of $x - 13$.
This gives me
\[\frac{1}{x + 4} \le 0.\]
Now, this is easy to solve: the only way for a fraction to be negative is if the denominator is negative.
So, I need $x + 4 < 0$, or $x < -4$.
But, I also have to remember that I assumed $x \ge 13$ at the beginning, so I need to take the intersection of these two conditions.
The intersection is empty, since there is no value of $x$ that satisfies both $x \ge 13$ and $x < -4$.
This means that the first case does not yield any solutions.
For the second case, I can assume that $x - 13 < 0$, or equivalently, $x < 13$.
Then, the inequality becomes
\[\frac{-(x - 13)}{x^2 - 9x - 52} \le 0.\]
Again, I can factor the denominator as $(x - 13)(x + 4)$, and this time I get
\[\frac{-1}{x + 4} \le 0.\]
Now, I need the fraction to be negative or zero, which means the denominator has to be positive or zero.
So, I need $x + 4 \ge 0$, or $x \ge -4$.
But, I also have to remember that I assumed $x < 13$, so I need to take the intersection of these two conditions.
The intersection is $-4 \le x < 13$, which is the solution for the second case.
Since the first case had no solutions, the final answer is just the solution for the second case.