2. Solve the inequality $\quad \frac{x^{2}}{3}+\frac{40}{x}+\frac{48}{x^{2}} \geq \frac{10|x|}{3}$.

(12 points)
Solution. 1) $x>0, \frac{x^{2}}{3}+\frac{40}{x}+\frac{48}{x^{2}} \geq \frac{10 x}{3}, \quad x^{2}+\frac{144}{x^{2}} \geq 10\left(x-\frac{12}{x}\right) . \quad$ Let's make the substitution $\quad t=x-\frac{12}{x}$, $t^{2}-10 t+24 \geq 0, \quad t \leq 4$, or $t \geq 6$. Returning to the variable $x: x-\frac{12}{x} \leq 4, \frac{x^{2}-4 x-12}{x} \leq 0$, $\frac{(x-6)(x+2)}{x} \leq 0 . \quad$ Considering $\quad x>0$, we have $\quad x \in(0 ; 6] . \quad$ Solving the second inequality $x-\frac{12}{x} \geq 6, \frac{x^{2}-6 x-12}{x} \geq 0, \frac{(x-3-\sqrt{21})(x-3+\sqrt{21})}{x} \leq 0$. Considering $x>0$, we have $x \in[3+\sqrt{21} ;+\infty)$. Therefore, $x \in(0 ; 6] \cup[3+\sqrt{21} ;+\infty)$.

2) $x<0, \frac{x^{2}}{3}+\frac{40}{x}+\frac{48}{x^{2}} \geq-\frac{10 x}{3}, \quad x^{2}+\frac{144}{x^{2}} \geq-10\left(x+\frac{12}{x}\right)$. Let's make the substitution $t=x+\frac{12}{x}, t^{2}+10 t-24 \geq 0$, $t \leq-12$, or $t \geq 2$. Returning to the variable $x: \quad x+\frac{12}{x} \leq-12, \frac{x^{2}+12 x+12}{x} \leq 0$, $\frac{(x-6-2 \sqrt{6})(x-6+2 \sqrt{6})}{x} \leq 0$. Considering $x<0$, we have $x \in(-\infty ;-6-2 \sqrt{6}] \cup[-6+2 \sqrt{6} ; 0) . \quad$ Solving the second inequality $x+\frac{12}{x} \geq 2, \frac{x^{2}-2 x+12}{x} \geq 0$. The numerator is positive for any value of $x$. The inequality has no negative solutions. Therefore, $x \in(-\infty ;-6-2 \sqrt{6}] \cup[-6+2 \sqrt{6} ; 0)$.

Answer: $x \in(-\infty ;-6-2 \sqrt{6}] \cup[-6+2 \sqrt{6} ; 0) \cup(0 ; 6] \cup[3+\sqrt{21} ;+\infty)$.