Решебник по алгебре 9 класс Мерзляк Задание 261

 

\[\boxed{\text{261\ (261).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \frac{1}{3}x + 12\]

\[\frac{1}{3}x + 12 = 0\]

\[\frac{1}{3}x = - 12\]

\[x = - 36.\ \]

\[Ответ:x = - 36\]

\[2)\ f(x) = 6x^{2} + 5x + 1\]

\[6x^{2} + 5x + 1 = 0\]

\[D = 25 - 24 = 1\]

\[x = \frac{- 5 \pm 1}{12}\]

\[x = - \frac{1}{2};\ \ \ x = - \frac{1}{3}.\]

\[Ответ:x = - \frac{1}{3};\ \ x = - 0,5.\]

\[3)\ f(x) = \sqrt{x^{2} - 4}\]

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

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Ответ:x = \pm 2.\]

\[4)\ f(x) = - 5\]

\[Ответ:\ нулей\ функции\ нет.\]

\[5)\ f(x) = \frac{3 - 0,2x}{x + 1};\ \ x \neq - 1\]

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

\[0,2x = 3\]

\[x = 15.\]

\[Ответ:x = 15.\]

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

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

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

\[x = 0;\ \ x = 1.\]

\(Ответ:x = 0;\ \ x = 1.\)

\[\boxed{\mathbf{261.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (2 - y)(3 + y) \leq (4 + y)(6 - y)\]

\[6 - 3y + 2y - y^{2} \leq 24 + 6y - 4y + y^{2}\]

\[- y - 2y \leq 24 - 6\]

\[- 3y \leq 18\]

\[y \geq - 6.\]

\[2)\ \frac{y - 1}{2} - \frac{2y + 1}{8} - y < 2\ \ \ | \cdot 8\]

\[4(y - 1) - 2y - 1 - 8y < 16\]

\[4y - 4 - 10y - 1 < 16\]

\[- 6y < 21\]

\[y > - \frac{21}{6}\]

\[y > - \frac{7}{2}\]

\[y > - 3,5.\]