Решебник по алгебре 9 класс Макарычев Задание 312

 

\[\boxed{\text{312\ (312).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 3x^{2} + 40x + 10 < - x^{2} +\]

\[+ 11x + 3\]

\[4x^{2} + 29x + 7 < 0\]

\[D = 29^{2} - 4 \cdot 4 \cdot 7 = 729\]

\[x_{1,2} = \frac{- 29 \pm 27}{8};\ \ \]

\[x_{1} = - \frac{1}{4};\ \ x_{2} = - 7.\]

\[4 \cdot (x + 0,25)(x + 7) < 0\]

\[x \in ( - 7;\ - 0,25).\]

\[\textbf{б)}\ 9x^{2} - x + 9 \geq 3x^{2} + 18x - 6\]

\[6x^{2} - 19x + 15 \geq 0\]

\[D = 19^{2} - 4 \cdot 6 \cdot 15 = 1\]

\[x_{1,2} = \frac{19 \pm 1}{12} = 1,5;1\frac{2}{3};\]

\[6 \cdot (x - 1,5)\left( x - 1\frac{2}{3} \right) \geq 0\]

\[x \in ( - \infty;1,5) \cup \left( 1\frac{2}{3};\ + \infty \right).\]

\[\textbf{в)}\ 2x^{2} + 8x -\]

\[- 111 < (3x - 5)(2x + 6)\]

\[2x^{2} + 8x - 111 < 6x^{2} + 18x -\]

\[- 10x - 30\]

\[4x^{2} + 81 > 0 \Longrightarrow\]

\[\Longrightarrow x - любое\ число.\]

\[\textbf{г)}\ (5x + 1)(3x - 1) >\]

\[> (4x - 1)(x + 2)\]

\[15x^{2} - 5x + 3x - 1 >\]

\[> 4x^{2} + 8x - x - 2\]

\[11x^{2} - 9x + 1 > 0\]

\[D = 81 - 44 = 37\]

\[x_{1,2} = \frac{9 \pm \sqrt{37}}{22}.\]

\[x \in \left( - \infty;\ \frac{9 - \sqrt{37}}{22} \right) \cup \left( \frac{9 + \sqrt{37}}{22}; + \infty \right).\]

\[\boxed{\text{312.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[1)\ x^{2} - 4x - 12 = 0\]

\[D_{1} = 4 + 12 = 16\]

\[x_{1} = 2 + 4 = 6;\ \ x_{2} = 2 - 4 =\]

\[= - 2.\]

\[2)\ x^{2} - 10x + 24 = 0\]

\[D_{1} = 25 - 24 = 1\]

\[x_{3} = 5 + 1 = 6;\ \ \ x_{4} = 5 - 1 = 4.\]

\[Ответ:x = 6.\]

\[1)\ x^{2} + 15x + 50 = 0\]

\[x_{1} + x_{2} = - 15;\ \ \ x_{1} \cdot x_{2} = 50\]

\[x_{1} = - 5;\ \ \ x_{2} = - 10.\]

\[2)\ x^{2} + 7x + 10 = 0\]

\[x_{1} + x_{2} = - 7;\ \ \ x_{1} \cdot x_{2} = 10\]

\[x_{1} = - 5;\ \ \ x_{2} = - 2.\]

\[Ответ:x = - 5.\]