16. Given that the function $f(x)$ is monotonically increasing on $[2,+\infty)$, and for any real number $x$ it always holds that $f(2+x)=f(2-x)$, if $f\left(1-2 x^{2}\right)<f\left(1+2 x-x^{2}\right)$, then the range of values for $x$ is
16. $(-2,0)$ From $f(2+x)=f(2-x)$, we know that the graph of $f(x)$ is symmetric about $x=2$. Combining this with the fact that $f(x)$ is monotonically increasing on $[2,+\infty)$, we can conclude that $f(x)$ is monotonically decreasing on $(-\infty, 2]$. Also,
$$
1-2 x^{2} \leqslant 1,1+2 x-x^{2}=2-(x-1)^{2} \leqslant 2,
$$

Therefore,
$$
\begin{array}{c}
f\left(1-2 x^{2}\right)1+2 x-x^{2},
\end{array}
$$

which means $x^{2}+2 x<0$, solving this gives $-2<x<0$.