Решебник по алгебре 9 класс Макарычев Задание 308

 

\[\boxed{\text{308\ (308).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} < 16\]

\[x^{2} - 16 < 0\]

\[(x - 4)(x + 4) < 0\]

\[x \in ( - 4;4).\]

\[\textbf{б)}\ x^{2} \geq 3\]

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

\[\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right) \geq 0\]

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

\[\textbf{в)}\ 0,2x^{2} > 1,8\]

\[0,2x^{2} - 1,8 > 0\ \ \ \ \ \ \ |\ :0,2\]

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

\[(x - 3)(x + 3) > 0\]

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

\[\textbf{г)} - 5x^{2} \leq x\]

\[5x^{2} + x \geq 0\]

\[5x(x + 0,2) \geq 0\]

\[x \in ( - \infty; - 0,2\rbrack \cup \lbrack 0; + \infty).\]

\[\textbf{д)}\ 3x^{2} < - 2x\]

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

\[3x\left( x + \frac{2}{3} \right) < 0\]

\[x \in \left( - \frac{2}{3};0 \right).\]

\[\textbf{е)}\ 7x < x^{2}\]

\[x^{2} - 7x > 0\]

\[x(x - 7) > 0\]

\[x \in ( - \infty;0) \cup (7; + \infty).\]

\[\boxed{\text{308.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{3} + 11x - 108 = 0\]

\[x_{1} = 4 \Longrightarrow корень.\ \]

\[Схема\ Горнера:\]

\[(x - 4)\left( x^{2} + 4x + 27 \right) = 0\]

\[x^{2} + 4x + 27 = 0\]

\[D = 4 - 27 < 0 \Longrightarrow корней\ нет.\]

\[\textbf{б)}\ x^{5} + 6x + 44 = 0\]

\[x_{1} = - 2 \Longrightarrow корень\ уравнения.\]

\[Схема\ Горнера:\]

\[(x + 2)\left( x^{4} - 2x^{3} + 4x^{2} - 8x + 22 \right) =\]

\[= 0\]

\[x^{4} - 2x^{3} + 4x^{2} - 8x + 22 =\]

\[= \left( x - \frac{1}{2} \right)^{4} + 2,5x^{2} - 7,5x +\]

\[+ 22 - \frac{1}{16} > 0\]

\[\Longrightarrow x = - 2.\]

\[Ответ:а)\ 4;\ \ б) - 2.\]