Решебник по алгебре 7 класс Мерзляк Задание 440

\[\boxed{\text{440\ (440).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ x^{2} - x = 0\]

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

\[x = 0\ \ или\ \ x - 1 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1.\]

\[Ответ:x = 0;1.\]

\[2)\ p^{2} + 15p = 0\]

\[p \cdot (p + 15) = 0\]

\[p = 0\ \ или\ \ p + 15 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p = - 15.\]

\[Ответ:p = 0;\ - 15.\]

\[3)\ 5x^{2} - 30x = 0\]

\[5x \cdot (x - 6) = 0\]

\[x = 0\ \ или\ \ x - 6 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 6\]

\[Ответ:x = 0;6.\]

\[4)\ 14x^{2} + 18x = 0\]

\[2x \cdot (7x + 9) = 0\]

\[x = 0\ \ или\ \ 7x + 9 = 0\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x = - \frac{9}{7} =\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 1\frac{2}{7}.\]

\[Ответ:\ x = 0;\ - 1\frac{2}{7}\text{.\ }\]