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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 427
Задание 427
\[\boxed{\text{427\ (427).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)y = \frac{5}{\sqrt{x^{2} - 4x - 12}} + \sqrt{x + 1}\]
\[\left\{ \begin{matrix} x^{2} - 4x - 12 > 0 \\ x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ x^{2} - 4x - 12 > 0\]
\(x_{1} + x_{2} = 4,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 6\)
\[x_{1}x_{2} = - 12,\ \ x_{2} = - 2\]
\[2)\ x + 1 \geq 0\]
\[x \geq - 1\]
\[\left\{ \begin{matrix} (x - 6)(x + 2) > 0 \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x \in (6;\ + \infty).\]
\[2)\ y = \frac{x - 3}{\sqrt{18 + 3x - x^{2}}} + \frac{8}{x - 5}\]
\[\left\{ \begin{matrix} 18 + 3x - x^{2} > 0 \\ x - 5 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ - x^{2} + 3x + 18 > 0\]
\(x_{1} + x_{2} = 3,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 6\)
\[x_{1}x_{2} = - 18,\ \ x_{2} = - 3\]
\[2)\ x - 5 \neq 0\]
\[x \neq 5\]
\[\left\{ \begin{matrix} (x - 6)(x + 3) > 0 \\ x \neq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x \in ( - 3;5) \cup (5;6).\]
\[3)\ y = \sqrt{x^{2} - 5x - 14} - \frac{9}{x^{2} - 81}\]
\[\left\{ \begin{matrix} x^{2} - 5x - 14 \geq 0 \\ x^{2} - 81 \neq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ x^{2} - 5x - 14 \geq 0\]
\(x_{1} + x_{2} = 5,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 7\)
\[x_{1}x_{2} = - 14,\ \ x_{2} = - 2\]
\[2)\ x^{2} - 81 \neq 0\]
\[x^{2} \neq 81\]
\[x \neq \pm 9\]
\[\left\{ \begin{matrix} (x - 7)(x + 2) \geq 0 \\ x \neq 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x \in ( - \infty; - 9) \cup\]
\[\cup ( - 9;\ - 2\rbrack \cup \lbrack 7;\ 9) \cup (9;\ + \infty).\]
\[4)\ y = \frac{1}{\sqrt{6 - 7x - 3x^{2}}} + \frac{2}{\sqrt{x + 1}}\]
\[\left\{ \begin{matrix} 6 - 7x - 3x^{2} > 0 \\ x + 1 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ - 3x^{2} - 7x + 6 > 0\]
\[D = 121\]
\[x_{1,2} = \frac{7 \pm 11}{- 6}\]
\[x = - 3;\ \ \ x = \frac{2}{3}\]
\[2)\ x + 1 > 0\]
\[x > - 1\]
\[\left\{ \begin{matrix} (x + 3)\left( x - \frac{2}{3} \right) > 0 \\ x > - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x \in \left( - 1;\frac{2}{3} \right)\text{.\ }\]
