Решебник по алгебре 9 класс Мерзляк Задание 329

 

\[\boxed{\text{329\ (329).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (x - 1)^{2} = \frac{2}{x};\ \ \ \ x \neq 0\]

\[y = (x - 1)^{2};\ \ y = \frac{2}{x}\]

\[Ответ:x = 2.\]

\[2)\ 1 - x^{2} = \sqrt{x} - 1;\ \ x \geq 0\]

\[y = 1 - x^{2};\ \ y = \sqrt{x} - 1\ \]

\[Ответ:x = 1.\]

\[\boxed{\text{329.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = 7x - 15\]

\[D(f) = ( - \infty; + \infty)\]

\[2)\ f(x) = \frac{8}{x + 5}\]

\[x + 5 \neq 0,\ \ x \neq - 5\]

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

\[3)\ f(x) = \frac{x - 10}{6}\]

\[D(f) = ( - \infty; + \infty)\]

\[4)\ f(x) = \sqrt{x - 9}\]

\[x - 9 \geq 0\]

\[x \geq 9\]

\[D(f) = \lbrack 9;\ + \infty)\]

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

\[1 - x > 0\]

\[x < 1\]

\[D(f) = ( - \infty;1)\ \]

\[6)\ f(x) = \frac{10}{x^{2} - 4}\]

\[x \neq 2,\ x \neq - 2\]

\[D(f) =\]

\[= ( - \infty;\ - 2) \cup ( - 2;2) \cup (2;\ + \infty)\]

\[7)\ f(x) = \frac{6x + 11}{x^{2} - 2x}\]

\[x^{2} - 2x \neq 0\]

\[x \cdot (x - 2) \neq 0\]

\[x \neq 0\]

\[x \neq 2\]

\[D(f) =\]

\[= ( - \infty;0) \cup (0;2) \cup (2;\ + \infty)\]

\[8)\ f(x) = \sqrt{x + 6} + \sqrt{4 - x}\]

\[x + 6 \geq 0\ \ \ и\ \ \ 4 - x \geq 0\]

\[\ \ \ \ \ \ \ x \geq - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 4\]

\[D(f) = \lbrack - 6;4\rbrack\text{.\ }\]