How many real numbers $x$ make $y=\frac{x^{2}-2 x+4}{x^{2}-3 x+3}$ an integer, and find the values of $x$ at this time.
Let $\frac{x+1}{x^{2}-3 x-3}=k$. ( $k$ is an integer)
There is $k x^{2}-(3 x+1) x+3 k-1=0$.
(1) When $k=0$, $-x-1=0 . x=-1$, which meets the requirement.
$(3 k+1)^{2}-4 k(3 k-1) \geqslant 0$.
Thus $3 k^{2}-10 k-1 \leqslant 0$.
We have $\frac{5-\sqrt{28}}{3} \leqslant k \leqslant \frac{5+\sqrt{28}}{3}$.
Since $k$ is an integer, hence $k=1,2,3$.
If $k=1, x^{2}-4 x+2=0, x=2 \pm \sqrt{2}$,
If $k=2, 2 x^{2}-7 x+5=0, x=1, x=\frac{5}{2}$,
If $k=3,3 x^{2}-10 x+8=0, x=2, x=\frac{4}{3}$.
Therefore, there are 7 values of $x$ that meet the requirement, $x_{1}=-1, x_{2}=2$
$$
\begin{array}{l}
+\sqrt{2}, x_{3}=2-\sqrt{2}, x_{4}=1, x_{5}=\frac{5}{2}, x_{5}=2, \\
x_{7}=\frac{4}{3} .
\end{array}
$$