[131 The equation $x^{2}-m x+2 m-2=0$ with respect to $x$ has a solution in the interval $\left[0, \frac{3}{2}\right]$, find the range of real number $m$.

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Analysis: We will divide the solutions in the interval $\left[0, \frac{3}{2}\right]$ into four cases: (1) a solution is 0; (2) a solution is $\frac{3}{2}$; (3) there is exactly one solution in the interval $\left(0, \frac{3}{2}\right)$; (4) there are two solutions in $\left(0, \frac{3}{2}\right)$. Then, we will combine the values of $m$ for each case.
Solution: Let $f(x)=x^{2}-m x+2 m-2$. According to the problem, we have the following cases:
(1) A solution is 0, then $f(0)=2 m-2=0$, so $m=1$;
(2) A solution is $\frac{3}{2}$, then $f\left(\frac{3}{2}\right)=\frac{9}{4}-\frac{3}{2} m+2 m-2=0$, so $m=-\frac{1}{2}$;
(3) There is exactly one solution in $\left(0, \frac{3}{2}\right)$, and 0 and $\frac{3}{2}$ are not solutions, then $f(0) \cdot f\left(\frac{3}{2}\right)=$ $(2 m-2)\left(\frac{1}{2} m+\frac{1}{4}\right)0 \\ f\left(\frac{3}{2}\right)>0,\end{array}\right.$
we get $\left\{\begin{array}{l}m \geqslant 4+2 \sqrt{2} \text { or } m \leqslant 4-2 \sqrt{2} \\ 01 \\ m>-\frac{1}{2},\end{array}\right.$
which means $1<m \leqslant 4-2 \sqrt{2}$.
In summary, the range of real numbers for $m$ is $\left[-\frac{1}{2}, 4-2 \sqrt{2}\right]$.