Решите неравенство: x^2-9x+8>0.

\[x^{2} - 9x + 8 > 0\]

\[D = ( - 9)^{2} - 4 \cdot 1 \cdot 8 =\]

\[= 81 - 32 = 49\]

\[x_{1} = \frac{9 + \sqrt{49}}{2 \cdot 1} = \frac{9 + 7}{2} = \frac{16}{2} = 8;\ \ \ \ \ \]

\[x_{2} = \frac{9 - \sqrt{49}}{2 \cdot 1} = \frac{9 - 7}{2} = \frac{2}{2} = 1\]

\[Ответ:( - \infty;1) \cup (8; + \infty).\]