ГДЗ по алгебре 9 класс Макарычев Задание 321

\[\boxed{\text{321\ (321).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = \sqrt{25 - x^{2}} + \sqrt{9x - x^{2} - 14}\]

\[ОДЗ:\]

\[\left\{ \begin{matrix} 25 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ 9x - x^{2} - 14 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ x^{2} - 9x + 14 \leq 0\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ (x - 2)(x - 7) \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 9x + 14 = 0\]

\[x_{1} + x_{2} = 9;\ \ \ x_{1} \cdot x_{2} = 14\]

\[x_{1} = 2;\ \ \ \ \ \ \ \ \ \ \ x_{2} = 7\]

\[\ x \in \lbrack 2;5\rbrack.\]

\[Ответ:\ x = 2;3;4;5.\]

\[\textbf{б)}\ y = \sqrt{8x - x^{2} - 12} +\]

\[+ \sqrt{16 - x^{2}}\]

\[ОДЗ:\]

\[\left\{ \begin{matrix} 8x - x^{2} - 12 \geq 0 \\ 16 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 8x + 12 \leq 0 \\ x^{2} - 16 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 2)(x - 6) \leq 0 \\ (x - 4)(x + 4) \leq 0 \\ \end{matrix} \right.\ \ \]

\[x^{2} - 8x + 12 = 0\]

\[D_{1} = 16 - 12 = 4\]

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

\[x_{2} = 4 - 2 = 2.\]

\[x \in \lbrack 2;4\rbrack.\]

\[Ответ:x = 2;3;4.\]