Решебник по алгебре 7 класс Мерзляк Задание 450

\[\boxed{\text{450\ (450).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[(x + 6) \cdot (2x - 9 - x) = 0\]

\[(x + 6) \cdot (x - 9) = 0\]

\[x + 6 = 0\ \ или\ \ x - 9 = 0\]

\[\ \ \ \ \ \ \ \ x = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 9.\]

\[Ответ:\ x = - 6;9.\]

\[(x - 10) \cdot (3x + 4 - x + 8) = 0\]

\[(x - 10) \cdot (2x + 12) = 0\]

\[x - 10 = 0\ \ или\ \ 2x + 12 = 0\]

\[\ \ \ \ \ \ \ \ \ \ x = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - 6.\]

\[Ответ:x = 10;\ - 6.\]

\[3)\ 3 \cdot (3x + 1)^{2} - 4 \cdot (3x + 1) =\]

\[= 0\]

\[(3x + 1) \cdot (9x + 3 - 4) = 0\]

\[(3x + 1) \cdot (9x - 1) = 0\]

\[3x + 1 = 0\ \ или\ \ 9x - 1 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ x = - \frac{1}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{1}{9}.\]

\[Ответ:\ x = - \frac{1}{3};\frac{1}{9}.\]

\[4)\ (9x - 12) - x \cdot (9x - 12) =\]

\[= 0\]

\[(9x - 12) \cdot (1 - x) = 0\]

\[9x - 12 = 0\ \ или\ \ 1 - x = 0\]

\[x = \frac{12}{9} = \frac{4}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1\]

\[x = 1\frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ .}\]

\[Ответ:x = 1\frac{1}{3};1.\ \]