Z13

По УРМ2
$$\left[\begin{array}{l}{x}^{2}-3x+2\ge 2x-{x}^{2}\\ {x}^{2}-3x+2\le {x}^{2}-2x\end{array}\right.\Rightarrow \left[\begin{array}{l}2{x}^{2}-5x+2\ge 0\\ -x+2\le 0\end{array}\right.\Rightarrow \left[\begin{array}{l}2{x}^{2}-5x+2\ge 0\\ x\ge 2\end{array}\right.\Rightarrow \left[\begin{array}{l}(x-0.5)(x-2)\ge 0\\ x\ge 2\end{array}\right.\Rightarrow \left[\begin{array}{l}x\in \left(-\infty ;0.5\right]\cup \left[2;\infty \right)\\ x\ge 2\end{array}\right.$$
Выходит $$x\in \left(-\infty ;0.5\right]\cup \left[2;\infty \right)$$


Ответ:
$$x\in \left(-\infty ;0.5\right]\cup \left[2;\infty \right)$$