\[\begin{array}{l}
Goi\,\,\,M\left( {{x_0};\,\,{y_0}} \right) \in dths\,\,\,y = \frac{{x - 1}}{{2x + 3}}\\
\Rightarrow M\left( {{x_0};\,\,\frac{{{x_0} - 1}}{{2{x_0} + 3}}} \right).\\
Diem\,\,\,M\,\,\,co\,\,toa\,\,\,do\,\,nguyen \Leftrightarrow \left\{ \begin{array}{l}
{x_0} \in Z\\
\,\frac{{{x_0} - 1}}{{2{x_0} + 3}} \in Z
\end{array} \right.\\
Ta\,\,co:\,\,\,\frac{{{x_0} - 1}}{{2{x_0} + 3}} \in Z \Rightarrow 2.\frac{{{x_0} - 1}}{{2{x_0} + 3}} \in Z\\
\Leftrightarrow \frac{{2{x_0} - 2}}{{2{x_0} + 3}} \in Z \Leftrightarrow \frac{{2{x_0} + 3 - 5}}{{2{x_0} + 3}} \in Z\\
\Leftrightarrow \left( {1 - \frac{5}{{2{x_0} + 3}}} \right) \in Z\\
\Rightarrow \frac{5}{{2{x_0} + 3}} \in Z\\
\Rightarrow 2{x_0} + 3 \in U\left( 5 \right)\\
\Rightarrow \left[ \begin{array}{l}
2{x_0} + 3 = - 5\\
2{x_0} + 3 = - 1\\
2{x_0} + 3 = 1\\
2{x_0} + 3 = 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x_0} = - 4\\
{x_0} = - 2\\
{x_0} = - 1\\
{x_0} = 1
\end{array} \right.\\
Thu\,\,\,lai\,\,ta\,\,duoc\,\,\,cac\,\,diem\,\,\,{M_1}\left( { - 4;\,1} \right),\,\,\,{M_2}\left( { - 2;\,3} \right),\\
{M_3}\left( { - 1; - 2} \right),\,\,{M_4}\left( {1;\,\,0} \right)\,\,\,thoa\,\,man.
\end{array}\]
