Đáp án:

$\begin{array}{l}
\left( {{m^2} - 4} \right).x + {m^2} - 3m + 2 = 0\\
 \Rightarrow \left( {{m^2} - 4} \right).x =  - \left( {{m^2} - 3m + 2} \right)\\
 \Rightarrow \left( {m - 2} \right)\left( {m + 2} \right).x =  - \left( {m - 1} \right)\left( {m - 2} \right)\left( * \right)\\
 + Khi:\left( {m - 2} \right)\left( {m + 2} \right) = 0 \Rightarrow \left[ \begin{array}{l}
m = 2\\
m =  - 2
\end{array} \right.\\
 + m = 2 \Rightarrow \left( * \right) \Rightarrow 0.x = 0\left( {\forall x} \right) \Rightarrow ktm\\
 + m =  - 2 \Rightarrow \left( * \right) \Rightarrow 0.x =  - 12\left( {ktm} \right)\\
 + Khi:m \ne 2;m \ne  - 2\\
\left( * \right) \Rightarrow x = \dfrac{{ - \left( {m - 1} \right)\left( {m - 2} \right)}}{{\left( {m - 2} \right)\left( {m + 2} \right)}} =  - \dfrac{{m - 1}}{{m + 2}}\\
 =  - \left( {\dfrac{{m + 2 - 3}}{{m + 2}}} \right)\\
 =  - \left( {1 - \dfrac{3}{{m + 2}}} \right)\\
 =  - 1 + \dfrac{3}{{m + 2}}\\
x \in Z\\
 \Rightarrow \dfrac{3}{{m + 2}} \in Z\\
 \Rightarrow \left( {m + 2} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow m \in \left\{ { - 5; - 3; - 1;1} \right\}\left( {tm} \right)\\
Vậy\,m \in \left\{ { - 5; - 3; - 1;1} \right\}
\end{array}$