Three, (25 points) Given $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$, and $y \leqslant|x|$. Find the range of values for $x$.


Three, when $x \geqslant 0$, the $x$ that meets the condition should satisfy $y-x \leqslant 0$, i.e., $\frac{1}{10}\left(x^{2}-11 x+18\right) \leqslant 0$. Since
$$
x^{2}-11x+18=(x-2)(x-3) \text {, }
$$

it is solved that $2 \leqslant x \leqslant 9$. Print $1+2 \leqslant x \leqslant 9$. Also, $\frac{1}{10}\left(x^{2}+9 x+18\right) \leqslant 0$. Since
$$
x^{2}+9 x+18=(x+3)(x+6) \text {, }
$$

it is solved that $-6 \leqslant x \leqslant-3$.
In summary, the range of $x$ is
$$
2 \leqslant x \leqslant 9 \text { or }-6 \leqslant x \leqslant-3 \text {. }
$$