4. Solve the inequality $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} \geq 0$. In the answer, indicate the sum of all natural numbers that are solutions to the inequality.
4. $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} * \frac{\left|x^{2}-3 x\right|+\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|+\left|x^{2}-2 x\right|} \geq 0 \Leftrightarrow \frac{\left(x^{2}-3 x\right)^{2}-\left(x^{2}-2\right)^{2}}{\left(x^{2}-x-2\right)^{2}-\left(x^{2}-2 x\right)^{2}} \geq 0 \Leftrightarrow$ $\Leftrightarrow \frac{\left(x^{2}-3 x-x^{2}+2\right)\left(x^{2}-3 x+x^{2}-2\right)}{\left(x^{2}-x-2-x^{2}+2 x\right)\left(x^{2}-x-2+x^{2}-2 x\right)} \geq 0 \Leftrightarrow \frac{(2-3 x)\left(2 x^{2}-3 x-2\right)}{(x-2)\left(2 x^{2}-3 x-2\right)} \geq 0 \Leftrightarrow$ $\frac{x-\frac{2}{3}}{x-2} \leq 0 \Rightarrow \frac{2}{3} \leq x<2$

. Answer: $\{1\}$