Решите систему неравенств x^2-8x+15<=0; 3x-13<=0.

Ответ:

\[\left\{ \begin{matrix} x^{2} - 8x + 15 \leq 0 \\ 3x - 13 \leq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - 8x + 15 \leq 0\]

\[D_{1} = 16 - 15 = 1\]

\[x_{1} = 4 + 1 = 5;\ \ x_{2} = 4 - 1 = 3.\]

\[(x - 3)(x - 5) \leq 0\]

\[3 \leq x \leq 5.\]

\[\left\{ \begin{matrix} 3 \leq x \leq 5 \\ 3x \leq 13\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} 3 \leq x \leq 5\ \ \ \ \ \\ x \leq \frac{13}{3} \leq 4\frac{1}{3} \\ \end{matrix} \right.\ \]

\[3 \leq x \leq 4\frac{1}{3}.\]

\[Ответ:3 \leq x \leq 4\frac{1}{3}.\]