Решебник по алгебре 9 класс Мерзляк Задание 260

 

\[\boxed{\text{260\ (260).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

\[0,2x = - 3\]

\[x = - 15.\]

\[Ответ:x = - 15.\]

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

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

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

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

\[Ответ:x = - 7;\ \ x = 5.\]

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

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

\[x + 3 = 0\]

\[x = - 3.\]

\[Ответ:x = - 3.\]

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

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

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

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

\[Ответ:x = - 2;\ \ x = 3.\]

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

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

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

\[x = 0;\]

\[x = \pm 2.\]

\[Ответ:x = \pm 2;x = 0.\]

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

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

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

\[нулей\ функции\ нет.\ \]

\[Ответ:нет\ нулей\ функции.\]

\[\boxed{\mathbf{260.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (4x - 3)^{2} + (3x + 2)^{2} \geq (5x + 1)^{2}\]

\[16x^{2} - 24x + 9 + 9x^{2} + 12x + 4 \geq 25x^{2} + 10x + 1\]

\[- 12x - 10x \geq 1 - 13\]

\[- 22x \geq - 12\]

\[x \leq \frac{12}{22}\]

\[x \leq \frac{6}{11}.\]

\[2)\ (6x - 1)^{2} - 4x(9x - 3) \leq 1\]

\[36x^{2} - 12x + 1 - 36x^{2} + 12x \leq 1\]

\[0x \leq 0\]

\[x - любое\ число.\]

\[3)\ \frac{2x - 1}{4} \geq \frac{3x - 5}{5}\text{\ \ \ \ }\]

\[5(2x - 1) \geq 4(3x - 5)\]

\[10x - 5 \geq 12x - 20\]

\[- 2x \geq - 15\]

\[x \leq 7,5.\]

\[4)\ \frac{x - 3}{9} - \frac{x + 4}{4} > \frac{x - 8}{6}\ \ \ | \cdot 36\]

\[4(x - 3) - 9(x + 4) > 6(x - 8)\]

\[4x - 12 - 9x - 36 > 6x - 48\]

\[- 3x - 6x > - 48 + 48\]

\[- 9x > 0\]

\[x < 0.\]