Найдите целые системы решений неравенств: x^2-x-2>0; 2x^2-13x+15<=0.

Ответ:

\[\left\{ \begin{matrix} x^{2} - x - 2 > 0\ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 13x + 15 \leq 0 \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - x - 2 > 0\]

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = 2;\ \ x_{2} = - 1.\]

\[(x + 1)(x - 2) > 0\]

\[x < - 1;\ \ x > 2.\]

\[2)\ 2x^{2} - 13x + 15 \leq 0\]

\[D = 169 - 120 = 49\]

\[x_{1} = \frac{13 + 7}{4} = 5;\ \ \]

\[x_{2} = \frac{13 - 4}{4} = \frac{6}{4} = 1,5.\]

\[(x - 1,5)(x - 5) \leq 0\]

\[1,5 \leq x \leq 5.\]

\[\left\{ \begin{matrix} x < - 1;\ \ x > 2 \\ 1,5 \leq x \leq 5\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2 < x \leq 5.\]

\[Ответ:2 < x \leq 5.\]