Ответ:
[-9; 9]
Объяснение:
0\\x^2\leq 81\end{cases}=> \begin{cases}x^2-2\cdot2,5x+6,25+0,75>0\\|x|\leq \sqrt{81}\end{cases}=> \\ \begin{cases}(x-2,5)^2+0,75>0\quad\quad (1)\\|x|\leq \sqrt{81} |x|\leq9\quad \:(2)\end{cases}=> ... \\" alt="\begin{cases}x^2-5x+7>0\\x^2\leq 81\end{cases}=> \begin{cases}x^2-2\cdot2,5x+6,25+0,75>0\\|x|\leq \sqrt{81}\end{cases}=> \\ \begin{cases}(x-2,5)^2+0,75>0\quad\quad (1)\\|x|\leq \sqrt{81} |x|\leq9\quad \:(2)\end{cases}=> ... \\" align="absmiddle" class="latex-formula">
(x-2,5)^2+0,75>0 \:\:\forall {x} \in R" alt="(x-2,5)^2\geq 0 => (x-2,5)^2+0,75>0 \:\:\forall {x} \in R" align="absmiddle" class="latex-formula">
(2)
\begin{cases}(x-2,5)^2+0,75>0\quad\quad (1)\\|x|\leq \sqrt{81} |x|\leq9\quad \:(2)\end{cases}=> \begin{cases}(x \in R, \\ x \in [-9;\: 9]\end{cases}=> x \in [-9;\: 9]" alt="...=>\begin{cases}(x-2,5)^2+0,75>0\quad\quad (1)\\|x|\leq \sqrt{81} |x|\leq9\quad \:(2)\end{cases}=> \begin{cases}(x \in R, \\ x \in [-9;\: 9]\end{cases}=> x \in [-9;\: 9]" align="absmiddle" class="latex-formula">