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МатематикаРешебник по алгебре 9 класс Макарычев Задание 393
Задание 393
\[\boxed{\text{393\ (393).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x - 8}{x + 4} > 0\]
\[(x + 4)(x - 8) > 0;\ \ \ x \neq - 4\]
\[x \in ( - \infty;\ - 4) \cup (8; + \infty).\]
\[\textbf{б)}\ \frac{x + 16}{x - 11} < 0\]
\[(x + 16)(x - 11) < 0;\ \ \ x \neq 11\]
\[x \in ( - 16;11).\]
\[\textbf{в)}\ \frac{x + 1}{3 - x} \geq 0\ \ \]
\[\left\{ \begin{matrix} (x + 1)(x - 3) \leq 0 \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in \lbrack - 1;3).\]
\[\textbf{г)}\ \frac{6 - x}{x - 4} \leq 0\]
\[\left\{ \begin{matrix} (x - 4)(x - 6) \geq 0 \\ x \neq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in ( - \infty;4) \cup \lbrack 6;\ + \infty).\]
\[\textbf{д)}\ \frac{2x - 4}{3x + 3} \leq 0\ \ \]
\[\left\{ \begin{matrix} (2x - 4)(3x + 3) \leq 0 \\ 3x \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2 \cdot (x + 1)(x - 2) \leq 0 \\ x \neq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in ( - 1;2\rbrack.\]
\[\textbf{е)}\ \frac{5x - 1}{2x + 3} \geq 0\ \ \]
\[\left\{ \begin{matrix} (5x - 1)(2x + 3) \geq 0\ \\ x \neq - 1,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 10 \cdot (x + 1,5)(x - 0,2) \geq 0\ \\ x \neq - 1,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in ( - \infty; - 1,5)\lbrack 0,2;\ + \infty).\]
\[\boxed{\text{393.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{2} + \frac{9}{x^{2}} - 10 = 0\ \ \ \ \ \ \ | \cdot x^{2}\]
\[x^{4} - 10x^{2} + 9 = 0\]
\[x^{2} = t:\]
\[t^{2} - 10t + 9 = 0\]
\[D_{1} = 25 - 9 = 16\]
\[t_{1} = 5 + 4 = 9;\ \ \ \ \ \]
\[\ t_{2} = 5 - 4 = 1.\]
\[1)\ x^{2} = 9\]
\[x_{1} = 3;\ \ \ \ x_{2} = - 3;\]
\[y_{1} = \frac{3}{3} = 1;\ \ \ \ y_{2} = \frac{3}{- 3} = - 1.\]
\[2)\ x^{2} = 1\]
\[x_{1} = 1;\ \ \ \ x_{2} = - 1;\]
\[y_{1} = \frac{3}{1} = 3;\ \ \ \ y_{2} = \frac{3}{- 1} = - 3.\]
\[Ответ:(3;1);( - 3; - 1);(1;3);\]
\[( - 1;\ - 3).\]
