Решите систему неравенств: -x^2+6x-8>=0; (3-x)/2>0.

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

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

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

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

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

\[\left\{ \begin{matrix}
2 \leq x \leq 4 \\
x < 3\ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \rightarrow 2 \leq x < 3.\]

\[Ответ:2 \leq x < 3.\]