ГДЗ по алгебре 9 класс Макарычев Задание 313

 

\[\boxed{\text{313\ (313).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x(3x - 1) > 4x^{2} + 5x + 9\]

\[6x^{2} - 2x > 4x^{2} + 5x + 9\]

\[2x^{2} - 7x - 9 > 0\]

\[D = 49 + 4 \cdot 2 \cdot 9 = 121\]

\[x_{1,2} = \frac{7 \pm 11}{4};\ \ \]

\[x_{1} = - 1;\ \ x_{2} = 4,5.\]

\[2 \cdot (x + 1)(x - 4,5) > 0\]

\[x \in ( - \infty;\ - 1) \cup (4,5;\ + \infty).\]

\[\textbf{б)}(5x + 7)(x - 2) < 21x^{2} -\]

\[- 11x - 13\]

\[5x^{2} - 10x + 7x - 14 < 21x^{2} -\]

\[- 11x - 13\]

\[16x^{2} - 8x + 1 > 0\]

\[(4x - 1)^{2} > 0\ \]

\[4x - 1 = 0\]

\[4x = 1\]

\[x = 0,25.\]

\[x \in ( - \infty;0,25) \cup (0,25;\ + \infty).\]

\[\boxed{\text{313.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{2} - 13x - \sqrt{5 - x} =\]

\[= - \sqrt{5 - x} - 30\]

\[x^{2} - 13x + 30 = 0\]

\[x_{1} + x_{2} = 13;\ \ \ x_{1} \cdot x_{2} = 30\]

\[Ответ:x = 3.\]