Question

помогите пожалуйста.​

Answer 1

$\displaystyle 8x-x^2\leq 0\\x(8-x)\leq 0\\\\\left \{ {{x\leq 0} \atop {8-x\geq 0}} \right. \rightarrow \left \{ {{x\leq 0} \atop {x\leq 8}} \right. \rightarrow x \in (- \infty, 0]\\ \\\left \{ {{x\geq 0} \atop {8-x\leq 0}} \right. \rightarrow \left \{ {{x\geq 0} \atop {x\geq 8}} \right. \rightarrow x \in [8, + \infty)\\ \\x \in (- \infty, 0] \cup[8, + \infty)$