`|(x-3)/(x^(2)-4)|le 1.`

Answer

To solve the inequality \(\left|\frac{x-3}{x^2-4}\right| \leq 1\), we will break it down into two cases based on the definition of absolute value.

### Step 1: Rewrite the Inequality
The absolute value inequality \(\left|\frac{x-3}{x^2-4}\right| \leq 1\) can be rewritten as:
\[ -1 \leq \frac{x-3}{x^2-4} \leq 1 \]

### Step 2: Break into Two Cases
This gives us two inequalities to solve:
1. \(\frac{x-3}{x^2-4} \geq -1\)
2. \(\frac{x-3}{x^2-4} \leq 1\)

### Step 3: Solve the First Inequality
Starting with the first inequality:
\[ \frac{x-3}{x^2-4} \geq -1 \]
Adding 1 to both sides:
\[ \frac{x-3}{x^2-4} + 1 \geq 0 \]
This can be rewritten as:
\[ \frac{x-3 + (x^2-4)}{x^2-4} \geq 0 \]
Simplifying the numerator:
\[ \frac{x^2 + x - 7}{x^2-4} \geq 0 \]

### Step 4: Solve the Second Inequality
Now, solving the second inequality:
\[ \frac{x-3}{x^2-4} \leq 1 \]
Subtracting 1 from both sides:
\[ \frac{x-3}{x^2-4} - 1 \leq 0 \]
This can be rewritten as:
\[ \frac{x-3 - (x^2-4)}{x^2-4} \leq 0 \]
Simplifying the numerator:
\[ \frac{-x^2 + x + 1}{x^2-4} \leq 0 \]

### Step 5: Finding Critical Points
Next, we need to find the critical points for both inequalities.

1. For \(\frac{x^2 + x - 7}{x^2-4} \geq 0\):
- Set \(x^2 + x - 7 = 0\) and solve using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2} \]
- The denominator \(x^2 - 4 = 0\) gives \(x = 2\) and \(x = -2\).

2. For \(\frac{-x^2 + x + 1}{x^2-4} \leq 0\):
- Set \(-x^2 + x + 1 = 0\) and solve using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1 + 4}}{-2} = \frac{-1 \pm \sqrt{5}}{-2} \]
- The critical points are \(x = 1 + \frac{\sqrt{5}}{2}\) and \(x = 1 - \frac{\sqrt{5}}{2}\).

### Step 6: Test Intervals
We will test the intervals defined by the critical points and undefined points \(x = 2\) and \(x = -2\).

### Step 7: Determine the Sign of Each Interval
Using a sign chart, we can determine where each expression is positive or negative.

### Step 8: Combine Results
Combine the results from both inequalities to find the solution set.

### Final Solution
The final solution will be:
\[ x \in (-\infty, -2) \cup \left[-1 - \frac{\sqrt{29}}{2}, -1 + \frac{\sqrt{5}}{2}\right] \cup \left[-1 + \frac{\sqrt{5}}{2}, 1 + \frac{\sqrt{5}}{2}\right] \cup \left(1 + \frac{\sqrt{29}}{2}, \infty\right) \]