Решите систему неравенств x^2+x-42<=0; 3x-5>0.

Ответ:

\[\left\{ \begin{matrix} x^{2} + x - 42 \leq 0 \\ 3x - 5 > 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + x - 42 \leq 0\]

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 42\]

\[x_{1} = - 7;\ \ \ x_{2} = 6.\]

\[(x + 7)(x - 6) \leq 0\]

\[- 7 \leq x \leq 6.\]

\[\left\{ \begin{matrix} - 7 \leq x \leq 6 \\ 3x > 5\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} - 7 \leq x \leq 6 \\ x > \frac{5}{3} > 1\frac{2}{3} \\ \end{matrix} \right.\ \]

\[1\frac{2}{3} < x \leq 6.\]

\[Ответ:1\frac{2}{3} < x \leq 6.\]