Решебник по алгебре 9 класс Мерзляк Задание 980

\[\boxed{\mathbf{980\ (980).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ y = \frac{4}{\sqrt{x^{2} + 3x - 10}} + \frac{1}{3x - 9}\]

\[\left\{ \begin{matrix} x^{2} + 3x - 10 > 0 \\ 3x - 9 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x + 5)(x - 2) > 0 \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\]

\[x \in ( - \infty; - 5) \cup (2;3) \cup (3; + \infty).\]

\[2)\ y = \frac{6}{\sqrt{12 + x - x²}} - \frac{2}{x² - 4}\]

\[\left\{ \begin{matrix} 12 + x - x^{2} > 0 \\ x^{2} - 4 \neq 0\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} - x - 12 < 0 \\ x \neq \pm 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x - 4)(x + 3) < 0 \\ x \neq \pm 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\]

\[x \in ( - 3; - 2) \cup ( - 2;2) \cup (2;4).\]

\[3)\ y = \sqrt{49 - x^{2}} + \frac{1}{\sqrt{x^{2} + 3x - 4}}\]

\[\left\{ \begin{matrix} 49 - x^{2} \geq 0\ \ \ \ \ \ \\ x^{2} + 3x - 4 > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} (7 - x)(7 + x) \geq 0 \\ (x + 4)(x - 1) > 0 \\ \end{matrix} \right.\ \]

\[Ответ:x \in \lbrack - 7;\ - 4) \cup (1;7\rbrack.\]