$9.251 \quad \frac{x^{2}-|x|-12}{x-3} \geq 2 x$.
Solution. The original inequality is equivalent to the inequality:

$\frac{x^{2}-6 x+|x|+12}{x-3} \leq 0$.

1) $\left\{\begin{array}{l}x \geq 0, \\ \frac{x^{2}-5 x+12}{x-3} \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ \frac{1}{x-3} \leq 0\end{array} \Leftrightarrow 0 \leq x<3\right.\right.$.
2) $\left\{\begin{array}{l}x<0, \\ \frac{(x-3)(x-4)}{x-3}, \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x<0, \\ x \leq 4, \\ x \neq 3\end{array} \Leftrightarrow x<0\right.\right.$.

Answer: $x \in(-\infty ; 3)$.