2. Solve the inequality $\frac{\sqrt[3]{\frac{x^{2}}{2}}-\sqrt[6]{4 x^{2}}}{\left(x^{2}-4|x|\right)^{2}-8 x^{2}+32|x|-48} \geqslant 0$.
Answer: $x \in(-\infty ;-6) \cup[-4 ;-2) \cup(-2 ; 2) \cup(2 ; 4] \cup(6 ;+\infty)$.

Solution. Consider the denominator of the fraction. It can be written as $\left(x^{2}-4|x|\right)^{2}-8\left(x^{2}-4|x|\right)-48$, or, if we let $x^{2}-4|x|=t$, in the form $t^{2}-8 t-48=(t-12)(t+4)$. If we return to the variable $x$, we get the expression $\left(x^{2}-4|x|-12\right)\left(x^{2}-4|x|+4\right)=(|x|-6)(|x|+2)(|x|-2)^{2}$. Thus, the original inequality is equivalent to the following

$$
\frac{\sqrt[3]{\frac{x^{2}}{2}}-\sqrt[3]{2|x|}}{(|x|-6)(|x|+2)(|x|-2)^{2}} \geqslant 0
$$

In the last inequality, we need to compare the fraction with zero, or, equivalently, determine the sign of this fraction. This means that if we replace the numerator or any of the factors in the denominator with an expression of the same sign, we will obtain an inequality equivalent to the original one. Note that the sign of the expression $\sqrt[3]{a}-\sqrt[3]{b}$ coincides with the sign of the expression $a-b$ for any $a$ and $b$; the expression $|a|-b$ for $b>0$ its sign coincides with the sign of the expression $a^{2}-b^{2}=(a-b)(a+b)$. Therefore, inequality (1) is equivalent to the following

$$
\frac{\frac{x^{2}}{2}-2|x|}{\left(x^{2}-36\right)\left(x^{2}-4\right)^{2}} \geqslant 0 \Leftrightarrow \frac{|x|(|x|-4)}{\left(x^{2}-36\right)\left(x^{2}-4\right)^{2}} \geqslant 0 \Leftrightarrow \frac{x^{2}(x-4)(x+4)}{(x-6)(x+6)(x-2)^{2}(x+2)^{2}} \geqslant 0
$$

Applying the method of intervals to the last inequality, we get

$x \in(-\infty ;-6) \cup[-4 ;-2) \cup(-2 ; 2) \cup(2 ; 4] \cup(6 ;+\infty)$.