Решите неравенство: x^2-5x+9>|x-6|.

\[x^{2} - 5x + 9 > |x - 6|\]

\[1)\ \left\{ \begin{matrix} x - 6 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x + 9 > x - 6 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x \geq 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 6x + 15 > 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 6x + 15 = 0\]

\[D = 36 - 60 < 0\]

\[\lbrack 6;\ + \infty).\]

\[2)\ \ \left\{ \begin{matrix} x - 6 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x + 9 > 6 - x \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x \leq 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4x + 3 > 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 4x + 3 = 0\]

\[x_{1} + x_{2} = 4\ \ \ \ \ \ \ x_{1} = 3\]

\[x_{1}x_{2} = 3\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 1\]

\[Ответ:( - \infty;\ \ 1) \cup (3;\ \ + \infty)\text{.\ }\]