$\frac{x^{4} + x^{2} + 1}{x^{2} - 4 x - 5} > 0$

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Solution

$\frac{x^{4} + x^{2} + 1}{x^{2} - 4 x - 5} > 0$
1) Numerator:
$x^{4} + x^{2} + 1$
$D \equiv 1 - 4 < 0, a > 0 \Rightarrow x^{4} + x^{2} + 1 > 0$

2)Denominator:
$x^{2} - 4 x - 5 > 0$
$\Rightarrow x^{2} - 5 x - x - 5 > 0$
$\Rightarrow (x + 1) (x - 5) > 0$
$x \in (- \infty, - 1) \cup (5, \infty)$

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\frac{x^{4} + x^{2} + 1}{x^{2} - 4 x - 5} > 0

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