Решебник по алгебре 9 класс Макарычев Задание 314

\[\boxed{\text{314\ (314).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = \sqrt{12x - 3x^{2}}\ \]

\[12x - 3x^{2} \geq 0\]

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

\[x(x - 4) \leq 0\]

\[x \in \lbrack 0;4\rbrack.\]

\[\textbf{б)}\ y = \frac{1}{\sqrt{2x^{2} - 12x + 18}}\]

\[2x^{2} - 12x + 18 > 0\]

\[x^{2} - 6x + 9 > 0\]

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

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

\[\boxed{\text{314.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{2} + 6x - 40 = 0\]

\[D_{1} = 9 + 40 = 49\]

\[Ответ:\ \ x = 4.\]