Решебник по алгебре 9 класс Мерзляк Задание 514

 

\[\boxed{\mathbf{514\ (514).\ }Еуроки - \ ДЗ\ без\ мороки}\]

\[1)\ (3x - 2)^{2} -\]

\[- (3x - 1)(2x + 3) < 3x(x - 7)\]

\[9x^{2} - 12x + 4 - 6x^{2} -\]

\[- 9x + 2x + 3 < 3x^{2} - 21x\]

\[3x^{2} - 21x + 2x + 7 -\]

\[- 3x^{2} + 21x < 0\]

\[2x + 7 < 0\]

\[x < - 3,5\]

\[Ответ:( - \infty;\ - 3,5).\]

\[2) - 3x^{2} - 10x + 48 \leq 0\]

\[3x^{2} + 10x - 48 \geq 0\]

\[D = 100 + 576 = 676\]

\[x_{1} = \frac{- 10 - 26}{6} = - 6\]

\[x_{2} = \frac{- 10 + 26}{6} = \frac{8}{3} = 2\frac{2}{3}\]

\[Ответ:\]

\[x \in ( - \infty;\ - 6\rbrack \cup \left\lbrack 2\frac{2}{3};\ + \infty \right).\]

\[\boxed{\text{514.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ |x| \cdot \left( x^{2} + 3x - 10 \right) < 0\]

\[\left\{ \begin{matrix} |x| > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3x - 10 < 0 \\ \end{matrix} \right.\ \]

\[|x| > 0\]

\[x^{2} + 3x - 10 < 0\]

\(x_{1} + x_{2} = - 3,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 5\)

\[x_{1}x_{2} = - 10,\ \ x_{2} = 2\]

\[Ответ:x \in ( - 5;0) \cup (0;2).\]

\[2)\ \sqrt{x} \cdot \left( x^{2} + 2x - 8 \right) \leq 0\]

\[\left\{ \begin{matrix} x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 8 \leq 0 \\ \end{matrix} \right.\ \]

\[x \geq 0\]

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

\(x_{1} + x_{2} = - 2,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 4\)

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

\[Ответ:x \in \lbrack 0;2\rbrack.\]

\[3)(x - 2)^{2}\left( x^{2} - 8x - 9 \right) < 0\]

\[\left\{ \begin{matrix} (x - 2)^{2} > 0\ \ \ \ \ \ \ \\ x^{2} - 8x - 9 < 0 \\ \end{matrix} \right.\ \]

\[1)\ (x - 2)^{2} > 0\]

\[x = 2\]

\[2)\ x^{2} - 8x - 9 < 0\]

\(x_{1} + x_{2} = 8,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 9\)

\[x_{1}x_{2} = - 9,\ \ x_{2} = - 1\]

\[Ответ:x \in ( - 1;2) \cup (2;9).\]

\[4)\ (x + 5)^{2}\left( x^{2} - 2x - 15 \right) > 0\]

\[\left\{ \begin{matrix} (x + 5)^{2} > 0\ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 15 > 0 \\ \end{matrix} \right.\ \]

\[1)\ (x + 5)^{2} > 0\]

\[x = - 5\]

\[2)\ x^{2} - 2x - 15 > 0\]

\(x_{1} + x_{2} = 2,\ \ \ \ \ \ \ \ \ \ \ x_{1} = 5\)

\[x_{1}x_{2} = - 15,\ \ x_{2} = - 3\]

\[Ответ:x \in ( - \infty; - 5) \cup\]

\[\cup ( - 5;\ - 3) \cup (5; + \infty).\]

\[5)\ \frac{x^{2} + 7x - 8}{(x - 4)^{2}} \geq 0\]

\[\left\{ \begin{matrix} x^{2} + 7x - 8 \geq 0 \\ (x - 4)^{2} > 0\ \ \ \ \ \ \\ x \neq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + 7x - 8 \geq 0\]

\(x_{1} + x_{2} = - 7,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 8\)

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

\[Ответ:x \in ( - \infty; - 8\rbrack \cup\]

\[\cup \lbrack 1;4) \cup (4;\ + \infty).\]

\[6)\ \frac{x^{2} + 10x - 11}{(x + 3)^{2}} \leq 0\]

\[\left\{ \begin{matrix} x^{2} + 10x - 11 \leq 0 \\ (x + 3)^{2} > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + 10x - 11 \leq 0\]

\(x_{1} + x_{2} = - 10,\ \ \ \ \ \ \ \ \ \ \ x_{1} = - 11\)

\[x_{1}x_{2} = - 11,\ \ x_{2} = 1\]

\[Ответ:x \in \lbrack - 11;\ - 3) \cup ( - 3;1\rbrack.\]