Ответ:

на фото ......................

Ответ: a∈(-∞;+∞).

Объяснение:

ОДЗ:

[tex]a^6+1\neq 0[/tex]

[tex]\displaystyle\\\frac{a^3}{1+a^6}\leq \frac{1}{2} \\\\\frac{a^3}{1+a^6}- \frac{1}{2} \leq 0\\\\\frac{2a^3-1*(1+a^6)}{2*(1+a^6)}\leq 0\\\\\frac{2a^3-1-a^6}{2(1+a^6)} \leq 0\\\\\frac{-(a^6-2a^3+1)}{2(a^6+1)}\leq 0\\\\\frac{-((a^3)^2-2*a^3*1+1^2)}{2(a^6+1)} \leq 0\\\\\frac{-(a^3-1)^2}{2(a^6+1)} \leq 0\ |*(-2)\\\\\frac{(a^3-1)^2}{a^6+1}\geq 0\\\\[/tex]

[tex]\\\\a^6\geq 0\\\\a^6+1\geq 0+1\\\\a^6+1\geq 1\ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ a\in(-\infty;+\infty).[/tex]

[tex](a^3-1)^2\geq 0\ \ \ \ \Rightarrow\ \ \ \ \ \ a\in(-\infty;+\infty).[/tex]