Решебник по алгебре 8 класс Мерзляк ФГОС Задание 627

\[\boxed{\mathbf{627\ (627).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² + 7x = 0\]

\[x(x + 7) = 0\]

\[x = 0,\ \ x = - 7\]

\[Ответ:\ x = - 7;0.\]

\[2)\ 2x² - 11x = 0\]

\[x(2x - 11) = 0\]

\[x = 0,\ \ 2x - 11 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 5,5\]

\[Ответ:x = 0;5,5.\]

\[3)\ 3x² - 6 = 0\]

\[3 \cdot \left( x^{2} - 2 \right) = 0\]

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

\[x^{2} = 2\]

\[x = \sqrt{2},\ \ x = - \sqrt{2}\]

\[Ответ:\ x = - \sqrt{2};\sqrt{2}.\]

\[4) - 8x^{2} = 0\]

\[x^{2} = 0\]

\[x = 0\]

\[Ответ:x = 0.\]

\[\boxed{\mathbf{627}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{x} \leq 8\ \ при\ x \leq 64;\]

\[2)\ \sqrt{x} > 7\ \ \ при\ x > 49;\]

\[3)\ 10 \leq \sqrt{x} \leq 20\ \ \ \ при\ \ \ \]

\[100 \leq x \leq 400.\]