Решите систему неравенств x^2-6x+8<=0; 3x-8>=0.

Ответ:

\[\left\{ \begin{matrix} x^{2} - 6x + 8 \leq 0 \\ 3x - 8 \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - 6x + 8 \leq 0\]

\[D_{1} = 9 - 8 = 1\]

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

\[(x - 2)(x - 4) \leq 0\]

\[2 \leq x \leq 4.\]

\[\left\{ \begin{matrix} 2 \leq x \leq 4 \\ 3x \geq 8\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} 2 \leq x \leq 4\ \ \ \ \\ x \geq \frac{8}{3} \geq 2\frac{2}{3} \\ \end{matrix} \right.\ \]

\[2\frac{2}{3} \leq x \leq 4.\]

\[Ответ:2\frac{2}{3} \leq x \leq 4.\]