2. Solve the inequality

$$
\sqrt{\frac{20-|x|}{x-3}}>x
$$
Answer. $x \in(-\infty ;-20] \cup(3 ; 4)$.

Solution. The domain of the inequality is determined by the condition $\frac{20-|x|}{x-3} \geqslant 0 \Leftrightarrow \frac{(20-x)(20+x)}{x-3} \geqslant 0$. Using the method of intervals, we get $x \in(-\infty ;-20] \cup(3 ; 20]$. On the interval $x \in(-\infty ;-20]$, the left side of the inequality is non-negative, while the right side is negative, so the inequality holds.

Consider the interval $x \in(3 ; 20]$. Then both sides of the inequality are non-negative, and they can be squared: $\frac{20-|x|}{x-3}>x^{2}$. Since the values of $x$ are greater than 3, the denominator of the fraction is positive, so we can multiply both sides of the inequality by it. Moreover, the absolute value can be dropped. We get $20-x > x^{2}(x-3) \Leftrightarrow x^{3}-3 x^{2}+x-20<0$. One of the roots of the polynomial on the left side is $x=4$. Factoring out the term $x-4$, the inequality becomes $(x-4)\left(x^{2}+x+5\right)<0$, from which $x<4$. Therefore, in this case, $3<x<4$ and finally $x \in(-\infty ;-20] \cup(3 ; 4)$.