To solve the inequality:

\[
\frac{|x^2 + 1|}{|x^2 - 5|} \leq 2
\]

### Step-by-step solution:

1. **Break down the inequality** into two cases based on the absolute values.

   Since \(x^2 + 1\) and \(x^2 - 5\) are quadratic expressions, they are always non-negative. This is because:
   - \(x^2 + 1 \geq 1\) for all real \(x\) (since \(x^2 \geq 0\)),
   - \(x^2 - 5 \geq 0\) only for \(x^2 \geq 5\) (i.e., \(x \geq \sqrt{5}\) or \(x \leq -\sqrt{5}\)).

   Therefore, the absolute values can be replaced by the expressions themselves depending on the sign of \(x^2 - 5\).

2. **Case 1: \(x^2 \geq 5\)**
   
   For \(x^2 \geq 5\), \(x \geq \sqrt{5}\) or \(x \leq -\sqrt{5}\), so the inequality simplifies to:

   \[
   \frac{x^2 + 1}{x^2 - 5} \leq 2
   \]

   Multiply both sides by \(x^2 - 5\) (which is positive in this region):

   \[
   x^2 + 1 \leq 2(x^2 - 5)
   \]

   Expand and simplify:

   \[
   x^2 + 1 \leq 2x^2 - 10
   \]

   \[
   1 + 10 \leq 2x^2 - x^2
   \]

   \[
   11 \leq x^2
   \]

   \[
   x \geq \sqrt{11} \quad \text{or} \quad x \leq -\sqrt{11}
   \]

3. **Case 2: \(x^2 < 5\)**

   For \(x^2 < 5\), the denominator \(x^2 - 5\) is negative, so the inequality becomes:

   \[
   \frac{x^2 + 1}{-(x^2 - 5)} \leq 2
  

Explore how to solve the inequality |x^2 + 1| / |x^2 - 5| ≤ 2 using key algebraic concepts. Learn about absolute values, quadratic expressions, and how to break down the inequality for different cases. Suitable for high school students in Grades 9-12.

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