Решебник по алгебре 9 класс Мерзляк Задание 342

\[\boxed{\text{342\ (342).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[f(x) = x^{2} - 2x - 15\]

\[1)\ x^{2} - 2x - 15 = 0\]

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 5\]

\[x_{1}x_{2} = - 15,\ \ x_{2} = - 3\]

\[Ответ:\ - 3;5.\]

\[2)\ x^{2} - 2x - 15 = - 7\]

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

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 4\]

\[x_{1}x_{2} = - 8,\ \ x_{2} = - 2\]

\[Ответ:4;\ - 2.\]

\[3)\ x^{2} - 2x - 15 = 33\]

\[x^{2} - 2x - 48 = 0\]

\[x_{1}{+ x}_{2} = 2,\ \ x_{1} = 8\]

\[x_{1}x_{2} = - 48,\ \ x_{2} = - 6\]

\[Ответ:8;\ - 6.\]

\[\boxed{\text{342.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ f(x) = \frac{x - 5}{(x - 2)(x - 1)}\]

\[2)\ f(x) = \frac{\sqrt{x - 5}}{x^{2} + 1}\]

\[3)\ f(x) = \frac{\sqrt{10 - x}}{5 \cdot (x + 1)}\]

\[4)\ f(x) = \sqrt{x + 4} + \sqrt{- 4 - x}\ \]