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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 431
Задание 431
\[\boxed{\text{431\ (431).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ |x| \cdot \left( x^{2} + 3x - 10 \right) < 0\]
\[\left\{ \begin{matrix} |x| > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3x - 10 < 0 \\ \end{matrix} \right.\ \]
\[|x| > 0\]
\[x^{2} + 3x - 10 < 0\]
\(x_{1} + x_{2} = - 3,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 5\)
\[x_{1}x_{2} = - 10,\ \ x_{2} = 2\]
\[Ответ:x \in ( - 5;0) \cup (0;2).\]
\[2)\ \sqrt{x} \cdot \left( x^{2} + 2x - 8 \right) \leq 0\]
\[\left\{ \begin{matrix} x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 8 \leq 0 \\ \end{matrix} \right.\ \]
\[x \geq 0\]
\[x^{2} + 2x - 8 \leq 0\]
\(x_{1} + x_{2} = - 2,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 4\)
\[x_{1}x_{2} = - 8,\ \ x_{2} = 2\]
\[Ответ:x \in \lbrack 0;2\rbrack.\]
\[3)(x - 2)^{2}\left( x^{2} - 8x - 9 \right) < 0\]
\[\left\{ \begin{matrix} (x - 2)^{2} > 0\ \ \ \ \ \ \ \\ x^{2} - 8x - 9 < 0 \\ \end{matrix} \right.\ \]
\[1)\ (x - 2)^{2} > 0\]
\[x = 2\]
\[2)\ x^{2} - 8x - 9 < 0\]
\(x_{1} + x_{2} = 8,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 9\)
\[x_{1}x_{2} = - 9,\ \ x_{2} = - 1\]
\[Ответ:x \in ( - 1;2) \cup (2;9).\]
\[4)\ (x + 5)^{2}\left( x^{2} - 2x - 15 \right) > 0\]
\[\left\{ \begin{matrix} (x + 5)^{2} > 0\ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 15 > 0 \\ \end{matrix} \right.\ \]
\[1)\ (x + 5)^{2} > 0\]
\[x = - 5\]
\[2)\ x^{2} - 2x - 15 > 0\]
\(x_{1} + x_{2} = 2,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 5\)
\[x_{1}x_{2} = - 15,\ \ x_{2} = - 3\]
\[Ответ:x \in ( - \infty; - 5) \cup\]
\[\cup ( - 5;\ - 3) \cup (5; + \infty).\]
\[5)\ \frac{x^{2} + 7x - 8}{(x - 4)^{2}} \geq 0\]
\[\left\{ \begin{matrix} x^{2} + 7x - 8 \geq 0 \\ (x - 4)^{2} > 0\ \ \ \ \ \ \\ x \neq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 7x - 8 \geq 0\]
\(x_{1} + x_{2} = - 7,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 8\)
\[x_{1}x_{2} = - 8,\ \ x_{2} = 1\]
\[Ответ:x \in ( - \infty; - 8\rbrack \cup\]
\[\cup \lbrack 1;4) \cup (4;\ + \infty).\]
\[6)\ \frac{x^{2} + 10x - 11}{(x + 3)^{2}} \leq 0\]
\[\left\{ \begin{matrix} x^{2} + 10x - 11 \leq 0 \\ (x + 3)^{2} > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 10x - 11 \leq 0\]
\(x_{1} + x_{2} = - 10,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 11\)
\[x_{1}x_{2} = - 11,\ \ x_{2} = 1\]
\[Ответ:x \in \lbrack - 11;\ - 3) \cup ( - 3;1\rbrack.\]
