Решебник по алгебре 9 класс Макарычев Задание 264

 

\[\boxed{\text{264}\text{\ (264)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = - \sqrt{x}\]

\[\textbf{б)}\ y = - \sqrt[3]{x}\]

\[\textbf{в)}\ y = \sqrt{- x}\]

\[\textbf{г)}\ y = \sqrt[3]{- x}\]

\[График\ функции\ y = - \sqrt{x}\ \]

\[симметричен\ графику\ \]

\[функции\ y = \sqrt{- x}\]

\[относительно\ начала\ \]

\[координат.\]

\[Графики\ функции\ y = - \sqrt[3]{x}\ и\ \ \]

\[y = \sqrt[3]{- x}\text{\ \ }совпадают.\]

\[\boxed{\text{264.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{2} + 2x - 48 < 0\]

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

\[D_{1} = 1 + 48 = 49\ \ \]

\[x_{1} = - 1 - 7 = - 8,\ \ \]

\[x_{2} = - 1 + 7 = 6\]

\[(x + 8)(x - 6) < 0\]

\[x \in ( - 8;6).\]

\[\textbf{б)}\ 2x^{2} - 7x + 6 > 0\]

\[2x^{2} - 7x + 6 = 0\]

\[D = 49 - 4 \cdot 2 \cdot 6 = 1\]

\[x_{1} = \frac{7 + 1}{4} = 2;\ \ \ x_{2} =\]

\[= \frac{7 - 1}{4} = 1,5;\]

\[2 \cdot (x - 2)(x - 1,5) > 0\]

\[x \in ( - \infty;1,5) \cup (2;\ + \infty).\]

\[\textbf{в)} - x^{2} + 2x + 15 < 0\]

\[x^{2} - 2x - 15 > 0\]

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

\[D_{1} = 1 + 15 = 16\]

\[x_{1} = 1 + 4 = 5;\ \ \ x_{2} =\]

\[= 1 - 4 = - 3.\]

\[(x - 5)(x + 3) > 0\]

\[x \in ( - \infty;\ - 3) \cup (5;\ + \infty).\]

\[\textbf{г)} - 5x^{2} + 11x - 6 > 0\]

\[5x^{2} - 11x + 6 < 0\]

\[5x^{2} - 11x + 6 = 0\]

\[D = 121 - 4 \cdot 5 \cdot 6 = 1\]

\[x_{1} = \frac{11 - 1}{10} = 1;\ \ \ \]

\[\ x_{2} = \frac{11 + 1}{10} = 1,2;\]

\[x \in (1;1,2).\]

\[\textbf{д)}\ 4x^{2} - 12x + 9 > 0\]

\[(2x - 3)^{2} > 0\]

\[2x - 3 = 0\]

\[2x = 3\]

\[x = 1,5.\]

\[x \in ( - \infty;1,5) \cup (1,5; + \infty).\]

\[\textbf{е)}\ 25x^{2} + 30x + 9 < 0\]

\[(5x + 3)^{2} < 0 \Longrightarrow решений\ нет.\]

\[\textbf{ж)} - 10x^{2} + 9x > 0\]

\[- 10x(x - 0,9) > 0\]

\[10x(x - 0,9) < 0\]

\[x = 0;\ \ x = 0,9.\]

\[x \in (0;0,9).\]

\[\textbf{з)} - 2x^{2} + 7x < 0\ \]

\[2x(x - 3,5) > 0\]

\[x = 0;\ \ \ x = 3,5.\]

\[x \in ( - \infty;0) \cup (3,5;\ + \infty).\]