Example 5. Solve the inequality

$$
|x-6|<x^{2}-5 x+9
$$
Solution. The given inequality is equivalent to the system

$$
\left\{\begin{array} { l } 
{ x - 6  - ( x ^ { 2 } - 5 x + 9 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}-6 x+15>0 \\
x^{2}-4 x+3>0
\end{array}\right.\right.
$$

The inequality $x^{2}-6 x+15>0$ holds for any $x$. Since $x^{2}-4 x+3=(x-1)(x-3)$, the inequality $x^{2}-4 x+3>0$ holds for $x<1$ or $x>3$.

Thus, the solution set of the original inequality (3) consists of the union of two intervals: $(-\infty ; 1) \cup(3 ;+\infty)$.

An inequality of the form

$$
|f(x)|>g(x)
$$

where $f(x)$ and $g(x)$ are some functions, is equivalent to the combination of two inequalities:

$$
f(x)>g(x), \quad f(x)<-g(x)
$$

An inequality of the form

$$
|f(x)|>a
$$

is equivalent to the combination

$$
\left[\begin{array}{l}
f(x)>a \\
f(x)<-a
\end{array}\right.
$$

If $a<0$, then the inequality of the form (5) holds for any admissible value of $x$ of the given inequality.