The inequality given is:

\[
\left|2 + \frac{5}{x}\right| > 1
\]

To solve this absolute value inequality, we need to break it into two cases based on the definition of absolute value.

### Case 1:
\[
2 + \frac{5}{x} > 1
\]

### Case 2:
\[
2 + \frac{5}{x} < -1
\]

#### Solving Case 1:
\[
2 + \frac{5}{x} > 1
\]
Subtract 2 from both sides:
\[
\frac{5}{x} > -1
\]
Take the reciprocal of both sides (note that taking the reciprocal reverses the inequality when the value is negative):
\[
x > -5
\]
So, for Case 1, \( x > -5 \).

#### Solving Case 2:
\[
2 + \frac{5}{x} < -1
\]
Subtract 2 from both sides:
\[
\frac{5}{x} < -3
\]
Take the reciprocal of both sides:
\[
x > -\frac{5}{3}
\]
So, for Case 2, \( x < -\frac{5}{3} \).

### Final Solution:
The solution is the intersection of the two cases:
- From Case 1: \( x > -5 \)
- From Case 2: \( x < -\frac{5}{3} \)

However, since \( x \) cannot be zero (as it would make the original expression undefined), we need to consider the intervals carefully.

### The intervals:
1. \( x < -\frac{5}{3} \) from Case 2.
2. \( -5 < x \) from Case 1.

The solution must satisfy both conditions simultaneously, which is possible in the interval \( x < -\frac{5}{3} \).

Therefore, the final solution is:

\[
x \in \left(-\infty, -\frac{5}{3}\right) \cup \left(0, 5\right)
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you solve absolute value inequalities with fractions?
2. What is the significance of the intersection of intervals in inequalities?
3. How can you verify the solution of an absolute value inequality?
4. What role does the undefined point \( x = 0 \) play in such problems?
5. How do you represent the solution of this inequality on a number line?

### Tip:
When solving absolute value inequalities, always remember to check for undefined points, as these can impact the solution intervals.

Learn how to solve the absolute value inequality |2 + (5)/(x)| > 1 with a detailed step-by-step solution. This problem covers inequalities involving absolute values and rational expressions. Suitable for students in Grades 10-12.

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