\[\begin{array}{l}
A = \left\{ {x \in Z|\,\,\frac{{6{x^2} - 5x + 5}}{{2x + 1}} \in Z} \right\}\\
M = \frac{{6{x^2} - 5x + 5}}{{2x + 1}} = 3x - 1 + \frac{6}{{2x + 1}}\\
\Rightarrow M \in Z \Leftrightarrow \frac{6}{{2x + 1}} \in Z\\
\Leftrightarrow 2x + 1 \in U\left( 6 \right) = \left\{ { \pm 1;\,\, \pm 2;\,\, \pm 3;\,\, \pm 6} \right\}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = - 6\\
2x + 1 = - 3\\
2x + 1 = - 2\\
2x + 1 = - 1\\
2x + 1 = 1\\
2x + 1 = 2\\
2x + 1 = 3\\
2x + 1 = 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \frac{7}{2}\,\,\left( {ktm} \right)\\
x = - 2\,\,\left( {tm} \right)\\
x = - \frac{3}{2}\,\,\left( {ktm} \right)\\
x = - 1\,\,\,\left( {tm} \right)\\
x = 0\,\,\,\left( {tm} \right)\\
x = \frac{1}{2}\,\,\left( {ktm} \right)\\
x = 1\,\,\,\left( {tm} \right)\\
x = \frac{5}{3}\,\,\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x \in \left\{ { - 2;\,\, - 1;\,\,0;\,\,1} \right\}.\\
\Rightarrow A = \left\{ { - 2;\,\, - 1;\,\,0;\,\,1} \right\}.\\
\Rightarrow {A_1} = \left\{ { - 2} \right\};\,\,\,{A_2} = \left\{ { - 1} \right\};\,\,\,{A_3} = \left\{ 0 \right\};\,\,{A_4} = \left\{ 1 \right\}.
\end{array}\]
