Решебник по алгебре 9 класс Мерзляк Задание 419

\[\boxed{\text{419\ (419).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ y = \sqrt{- x^{2} + 3x + 4}\]

\[- x^{2} + 3x + 4 \geq 0\]

\[x_{1} + x_{2} = 3,\ \ x_{1} = 4\]

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

\[Ответ:D(y) = \lbrack - 1;4\rbrack.\]

\[2)\ y = \sqrt{2x^{2} + 5x - 3}\]

\[2x^{2} + 5x - 3 \geq 0\]

\[D = 25 + 24 = 49\]

\[x_{1,2} = \frac{- 5 \pm 7}{4}\]

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

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

\[3)\ y = \frac{1}{\sqrt{x^{2} + 4x - 12}}\]

\[x^{2} + 4x - 12 > 0\]

\[x_{1} + x_{2} = - 4,\ \ x_{1} = - 6\]

\[x_{1}x_{2} = - 12,\ \ x_{2} = 2\]

\[Ответ:x \in ( - \infty;\ - 6) \cup (2;\ + \infty).\]

\[4)\ y = \frac{x + 2}{\sqrt{6x - 2x^{2}}}\]

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

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

\[x_{1} = 0,\ \ x_{2} = 3\]

\[Ответ:x \in (0;3).\]