Решебник по алгебре 9 класс Мерзляк Задание 662

 

\[\boxed{\mathbf{662\ (662).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ f(x) = \sqrt{3 - 5x - 2x^{2}}\]

\[3 - 5x - 2x^{2} \geq 0\]

\[2x^{2} + 5x - 3 \leq 0\]

\[D = 25 + 24 = 49\]

\[x_{1,\ 2} = \frac{- 5 \pm 7}{4}\]

\[x = - 3;\ \ x = 0,5\]

\[Ответ:x \in \lbrack - 3;0,5\rbrack\text{.\ }\]

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

\[3 - 5x - 2x^{2} > 0\]

\[2x² + 5x - 3 < 0\]

\[из\ (1) \Longrightarrow x_{1,2} = \frac{- 5 \pm 7}{4}\]

\[x = - 3;\ \ x = 0,5\]

\[Ответ:x \in ( - 3;0,5).\]

\[3)\ f(x) = \sqrt{3 - 5x - 2x^{2}} +\]

\[+ \frac{1}{x^{2} - 9}\]

\[\left\{ \begin{matrix} 3 - 5x - 2x^{2} \geq 0 \\ x^{2} - 9 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ,\ \ из\ (1)\]

\[Ответ:x \in ( - 3;0,5).\]

\[4)\ f(x) = \sqrt{3 - 5x - 2x^{2}} +\]

\[+ \frac{2}{x^{2} + 2x}\]

\[\left\{ \begin{matrix} 3 - 5x - 2x^{2} \geq 0,\ \ из\ (1)\ \ \ \ x \in \left\lbrack - 3;\frac{1}{2} \right\rbrack \\ x² + 2x \neq 0,\ \ x \neq 0,\ \ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[Ответ:x \in \lbrack - 3;\ - 2) \cup\]

\[\cup \ ( - 2;0) \cup (0;0,5\rbrack.\]

\[\boxed{\mathbf{662.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\frac{1}{216};\ \frac{1}{36};\ \frac{1}{6};\ldots\]

\[q = \frac{1}{36}\ :\frac{1}{216} = \frac{216}{36} = 6;\]

\[b_{5} = b_{1} \cdot q^{4} = \frac{1}{216} \cdot 6^{4} = \frac{6^{4}}{6^{3}} = 6.\]

\[Ответ:q = 6;\ \ \ b_{5} = 6.\]