Решебник по алгебре 9 класс Макарычев Задание 302

 

\[\boxed{\text{302\ (302).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = x^{2} - 3\]

\[x^{2} - 3 = 0\]

\[x^{2} = 3\]

\[x = \pm \sqrt{3}.\]

\[\textbf{а)}\ y > 0\ \ при\ \ \]

\[x \in \left( - \infty;\ - \sqrt{3} \right) \cup \left( \sqrt{3}; + \infty \right);\]

\[\textbf{б)}\ y < 0\ \ \ при\ \ x \in \left( - \sqrt{3};\sqrt{3} \right).\]

\[\boxed{\text{302.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{3} - 4x^{2} + 3x + 2 = 0;\ \ \]

\[x = 2 \Longrightarrow один\ из\ корней\ \]

\[уравнения.\]

\[(x - 2)\left( x^{2} - 2x - 1 \right) = 0\]

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

\[D_{1} = 1 + 1 = 2\]

\[x = 1 \pm \sqrt{2}.\]

\[Ответ:2;1 - \sqrt{2};1 + \sqrt{2}.\]

\[\textbf{б)}\ x^{4} + 2x^{3} - 7x^{2} - 8x + 12 =\]

\[= 0\ \ \]

\[x_{1} = 1 \Longrightarrow корень.\]

\[(x - 1)\left( x^{3} + 3x^{2} - 4x - 12 \right) =\]

\[= 0\ \ \]

\[x_{2} = 2 \Longrightarrow корень.\]

\[(x - 1)(x - 2)\left( x^{2} + 5x + 6 \right) =\]

\[= 0;\ \ \]

\[x^{2} + 5x + 6 = 0\]

\[x_{1} + x_{2} = - 5;\ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{3,4} = - 3;\ - 2.\]

\[Ответ:\ - 3;\ - 2;1;2.\]