ГДЗ по алгебре 8 класс Макарычев Задание 203

\[\boxed{\text{203\ (203).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \frac{x^{2} - 6x + 1}{x - 3} =\]

\[= \frac{x^{2} - 6x + 9 - 8}{x - 3} =\]

\[= \frac{(x - 3)^{2} - 8}{x - 3} =\]

\[= \frac{(x - 3)^{2}}{x - 3} - \frac{8}{x - 3} =\]

\[= x - 3 - \frac{8}{x - 3}\]

\[y \in Z;\ \ \ \ \frac{8}{x - 3} \in Z;\ \ \ \]

\[x - 3 \in \left\{ - 8, - 4, - 2,\ 1,\ 2,\ 4,\ 8 \right\}\text{.\ }\]

\[x = - 5\]

\[y = - 5 - 3 - \frac{8}{- 5 - 3} = - 7\]

\[x = - 1\]

\[y = - 1 - 3 - \frac{8}{- 1 - 3} = - 2\]

\[x = 1\]

\[y = 1 - 3 - \frac{8}{1 - 3} = 2\]

\[x = 2\]

\[y = 2 - 3 - \frac{8}{2 - 3} = 7\]

\[x = 4\]

\[y = 4 - 3 - \frac{8}{4 - 3} = - 7\]

\[x = 5\]

\[y = 5 - 3 - \frac{8}{5 - 3} = - 2\]

\[x = 7\]

\[y = 7 - 3 - \frac{8}{7 - 3} = 2\]

\[x = 11\]

\[y = 11 - 3 - \frac{8}{11 - 3} = 7\]

\[Ответ:( - 5;\ - 7),( - 1; - 2),\ (1;2),\ \]

\[(2;7),\ (4; - 7),\ (5; - 2),(7;2),\ \]

\[(11;7).\]