If $\left|{\mathrm{x}}^2-2\mathrm{x}+2\right|-\left|2{\mathrm{x}}^2-5\mathrm{x}+2\right|=\left|{\mathrm{x}}^2-3\mathrm{x}\right|$, then the set of values of x is

We have

$\left|2{\mathrm{x}}^2-5\mathrm{x}+2\right|+\left|{\mathrm{x}}^2-3\mathrm{x}\right|=\left|{\mathrm{x}}^2-2\mathrm{x}+2\right|$

$=\left|\left(2{\mathrm{x}}^2-5\mathrm{x}+2\right)-\left({\mathrm{x}}^2-3\mathrm{x}\right)\right|$

$\begin{array}{r}\therefore \left(2{\mathrm{x}}^2-5\mathrm{x}+2\right)\left({\mathrm{x}}^2-3\mathrm{x}\right)\le 0\\ \Rightarrow (2\mathrm{x}-1)(\mathrm{x}-2)\mathrm{x}(\mathrm{x}-3)\le 0\end{array}$

From the sign scheme, we have $\mathrm{x}\in \left[0,\frac{1}{2}\right]\cup [2,3]$