Решебник по алгебре 9 класс Мерзляк Задание 330

 

\[\boxed{\text{330\ (330).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{3}{x} = \sqrt{x} + 2;\ x \neq 0;\ \ x \geq 0\]

\[y = \frac{3}{x}\]

\[y = \sqrt{x} + 2\]

\[1)\ y = \sqrt{x}\]

\[2)\ y = \sqrt{x} + 2\]

\[Ответ:x = 1.\]

\[\boxed{\text{330.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \frac{x + 3}{x - 4}\]

\[x - 4 \neq 0\]

\[x \neq 4\]

\[D(f) = ( - \infty;4) \cup (4;\ + \infty).\]

\[2)\ f(x) = \frac{9}{x^{2} + 16}\]

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

\[3)\ f(x) = \frac{5x + 1}{x^{2} - 6x + 8}\]

\[x^{2} - 6x + 8 \neq 0\]

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

\[x_{1}x_{2} = 8,\ \ x_{2} = 4\]

\[D(f) =\]

\[= ( - \infty;2) \cup (2;4) \cup (4;\ + \infty).\]

\[4)\ f(x) = \sqrt{x - 1} + \sqrt{x - 3}\]

\[x - 1 \geq 0\ \ \ и\ \ \ x - 3 \geq 0\]

\[\ \ \ \ \ \ \ \ x \geq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \geq 3\]

\[D(f) = \lbrack 3; + \infty).\]

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

\[x - 5 \geq 0\ \ \ и\ \ \ 5 - x \geq 0\]

\[\ \ \ \ \ \ \ \ x \geq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 5\]

\[D(f) = \left\{ 5 \right\}\]

\[6)\ f(x) = \sqrt{x^{2} + 1}\]

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