Решите систему неравенств: x^2-2x-80<=0; x^2-2x-24>0.

Ответ:

\[\left\{ \begin{matrix} x² - 2x - 80 \leq 0 \\ x² - 2x - 24 > 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 2x - 80 = 0\]

\[x_{1} + x_{2} = 2,\ \ x_{1} = - 80\]

\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = 10\]

\[\lbrack - 8;10\rbrack.\]

\[x² - 2x - 24 > 0\]

\[x_{1} + x_{2} = 2,\ \ x_{1} = - 24\]

\[x_{1} \cdot x_{2} = 6,\ \ x_{2} = - 4\]

\[( - \infty;\ - 4)(6;\ + \infty).\]

\[Ответ:\lbrack - 8; - 4) \cup (6;10\rbrack.\]