Решите неравенство: корень из (2x-8)<=2.

\[\sqrt{2x - 8} \leq 2\]

\[t = 2x - 8\]

\[\sqrt{t} \leq 2\]

\[0 \leq t \leq 4 \Longrightarrow 0 \leq 2x - 8 \leq 4\]

\[\left\{ \begin{matrix} 2x - 8 \geq 0 \\ 2x - 8 \leq 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x \geq 8\ \ \ \\ 2x \leq 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x \geq 4 \\ x \leq 6 \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack 4;6\rbrack\text{.\ }\]