2. Solve the inequality

$$
\sqrt{\frac{|x|-12}{2-x}}>x
$$
Answer. $x \in(-\infty ;-12] \cup(2 ; 3)$.

Solution. The domain of the inequality is determined by the condition $\frac{|x|-12}{2-x} \geqslant 0 \Leftrightarrow \frac{(x-12)(x+12)}{2-x} \geqslant 0$. Using the method of intervals, we get $x \in(-\infty ;-12] \cup(2 ; 12]$. On the interval $x \in(-\infty ;-12]$, the left side of the inequality is non-negative, while the right side is negative, so the inequality holds.

Consider the interval $x \in(2 ; 12]$. Then both sides of the inequality are non-negative, and they can be squared: $\frac{|x|-12}{2-x}>x^{2}$. Since the values of $x$ are greater than 2, the denominator of the fraction is negative - we can multiply both sides of the inequality by it and change the sign. Moreover, the absolute value can be dropped. We get $x-12<x^{2}(2-x) \Leftrightarrow x^{3}-2 x^{2}+x-12<0$. One of the roots of the polynomial on the left side is $x=3$. Factoring out the term $x-3$, the inequality becomes $(x-3)\left(x^{2}+x+4\right)<0$, from which $x<3$. Therefore, in this case $2<x<3$ and finally $x \in(-\infty ;-12] \cup(2 ; 3)$.