To solve the inequality \(\frac{|2x - 1|}{|x + 5|} \leq 0\), let's analyze it step by step.

### Step 1: Consider the nature of absolute values
- The absolute value \(|2x - 1|\) is always non-negative, i.e., \(|2x - 1| \geq 0\).
- The absolute value \(|x + 5|\) is also always non-negative, i.e., \(|x + 5| \geq 0\).

### Step 2: Analyze the inequality \(\frac{|2x - 1|}{|x + 5|} \leq 0\)
- A fraction is less than or equal to zero only when the numerator is zero and the denominator is non-zero.

### Step 3: Set the numerator to zero
\[
|2x - 1| = 0
\]
\[
2x - 1 = 0 \implies x = \frac{1}{2}
\]

### Step 4: Ensure the denominator is non-zero at \(x = \frac{1}{2}\)
\[
|x + 5| \neq 0 \implies x + 5 \neq 0 \implies x \neq -5
\]

### Step 5: Combine the results
- The inequality \(\frac{|2x - 1|}{|x + 5|} \leq 0\) holds only when \(x = \frac{1}{2}\), since this makes the numerator zero and the entire expression zero.

### Final Answer:
\[
x = \frac{1}{2}
\]

This is the only solution.

Would you like more details or have any questions? Here are five related questions for further exploration:

1. What happens if the inequality is strictly less than zero, \(\frac{|2x - 1|}{|x + 5|} < 0\)?
2. How would the solution change if the absolute values were not present?
3. What if we consider a similar inequality but with a different function, like \(\frac{|x^2 - 4|}{|x + 5|} \leq 0\)?
4. Can you solve the inequality \(\frac{|2x - 1|}{x + 5} \leq 0\) without absolute value in the denominator?
5. How does the solution process differ if the inequality is non-strict, such as \(\frac{|2x - 1|}{|x + 5|} \geq 0\)?

**Tip:** When working with absolute values in inequalities, carefully consider the cases where the expression inside the absolute value is zero, as this often defines critical points for the solution.

Explore how to solve the absolute value inequality |2x-1| / |x+5| ≤ 0 with a comprehensive, step-by-step guide. This problem involves key algebraic concepts such as inequalities and absolute values, making it ideal for students in Grades 10-12.

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