Решебник по алгебре 9 класс Мерзляк Задание 356

 

\[\boxed{\text{356\ (356).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = 2x^{2} - 3x + 6;\ \ \ y = x + 12\]

\[- 2x^{2} - 3x + 6 = x + 12\]

\[2x^{2} - 4x - 6 = 0\ \ |\ :2\]

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

\[x_{1} + x_{2} = 2,\ \ x_{1}x_{2} = - 3,\ \ \]

\[x_{1} = - 1,\ \ x_{2} = 3.\]

\[x = - 1 \rightarrow \ \ y = 11\]

\[x = 3 \rightarrow \ y = 15.\]

\[Ответ:( - 1;11),\ (3;15).\]

\[\boxed{\text{356.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = 0,2x + 3\]

\[0,2x + 3 = 0\]

\[0,2x = - 3\]

\[x = - 15.\]

\[Ответ:x = - 15.\]

\[2)\ g(x) = 35 - 2x - x^{2}\]

\[x^{2} + 2x - 35 = 0\]

\[x_{1} + x_{2} = - 2,\ \ x_{1} = - 7\]

\[x_{1}x_{2} = - 35,\ \ x_{2} = 5.\]

\[Ответ:x = - 7;\ \ x = 5.\]

\[3)\ \varphi(x) = \sqrt{x + 3}\]

\[\sqrt{x + 3} = 0\]

\[x + 3 = 0\]

\[x = - 3.\]

\[Ответ:x = - 3.\]

\[4)\ h(x) = \frac{x^{2} - x - 6}{x + 3};\ \ x \neq - 3\]

\[x^{2} - x - 6 = 0\]

\[x_{1} + x_{2} = 1,\ \ x_{1} = 3\]

\[x_{1}x_{2} = - 6,\ \ x_{2} = - 2\]

\[Ответ:x = - 2;\ \ x = 3.\]

\[5)\ f(x) = x^{3} - 4x\]

\[x^{3} - 4x = 0\]

\[x \cdot \left( x^{2} - 4 \right) = 0\]

\[x = 0;\]

\[x = \pm 2.\]

\[Ответ:x = \pm 2;x = 0.\]

\[6)\ f(x) = x^{2} + 1\]

\[x^{2} + 1 = 0\]

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

\[нулей\ функции\ нет.\ \]

\[Ответ:нет\ нулей\ функции.\]