Решебник по алгебре 9 класс Мерзляк Задание 242

\[\boxed{\text{242\ (242).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \sqrt{x - 2} + \frac{x + 2}{x - 5}\]

\[\left\{ \begin{matrix} x - 2 \geq 0 \\ x - 5 \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 2 \\ x \neq 5 \\ \end{matrix} \right.\ \]

\[D(f) = \lbrack 2;5) \cup (5;\ + \infty)\ \]

\[Ответ:D(f) = \lbrack 2;5) \cup (5;\ + \infty).\]

\[2)\ f(x) = \frac{x}{|x| - 7}\]

\[|x| - 7 \neq 0\]

\[\left\{ \begin{matrix} x - 7 \neq 0 \\ - x - 7 \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \neq 7 \\ x \neq - 7 \\ \end{matrix} \right.\ \]

\[D(f) =\]

\[= ( - \infty;\ - 7) \cup ( - 7;7) \cup (7;\ + \infty)\text{.\ }\]

\[Ответ:D(f) =\]

\[= ( - \infty;\ - 7) \cup ( - 7;7) \cup (7;\ + \infty).\]

\[3)\ f(x) = \sqrt{x + 3} + \frac{1}{x^{2} - 9}\]

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

\[D(f) = ( - 3;3) \cup (3; + \infty)\text{.\ }\]

\[Ответ:D(f) =\]

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

\[4)\ f(x) = \frac{\sqrt{x - 4}}{\sqrt{x + 2}} + \frac{4x - 3}{x^{2} - 7x + 6}\]

\[\left\{ \begin{matrix} x - 4 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + 2 > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 7x + 6 \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 4\ \ \ \ \\ x > - 2\ \\ x \neq 1\ \ \ \ \\ x \neq 6\ \ \ \ \\ \end{matrix} \right.\ \]

\[x_{1} + x_{2} = 7,\ \ x_{1} = 1\]

\[x_{1}x_{2} = 6,\ \ x_{2} = 6\]

\[D(f) = \lbrack 4;6) \cup (6; + \infty)\text{.\ }\]

\[Ответ:D(f) = \lbrack 4;6) \cup (6; + \infty)\text{.\ }\]