tìm tọa độ các điểm nguyên thuộc đồ thị hám số y = (x-1)/( 2x+3))

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