Решебник по алгебре 9 класс Макарычев Задание 348

 

\[\boxed{\text{348\ (348).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[Схема\ Горнера:\]

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

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

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

\[\textbf{б)}\ x^{5} + 6x + 44 = 0\]

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

\[Схема\ Горнера:\]

\[(x + 2)\left( x^{4} - 2x^{3} + 4x^{2} - 8x + 22 \right) = 0\]

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

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

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

\[\Longrightarrow x = - 2.\]

\[Ответ:а)\ 4;\ \ б) - 2.\]

\[\boxed{\text{348.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + x - 6 < 0\ \ \ \\ - x^{2} + 2x + 3 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x - 6 < 0\ \ \ \ (1) \\ x^{2} - 2x - 3 < 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x^{2} + x - 6 = 0\ \ \]

\[D = 1 + 4 \cdot 6 = 25\ \ \]

\[x_{1,2} = \frac{- 1 \pm 5}{2} = - 3;2.\]

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

\[D_{1} = 1 + 3 = 4\]

\[x_{1,2} = 1 \pm 2 = - 1;3.\]

\[\left\{ \begin{matrix} (x + 3)(x - 2) < 0 \\ (x + 1)(x - 3) < 0 \\ \end{matrix} \right.\ \]

\[x \in ( - 1;2).\]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + 4x - 5 > 0\ \ (1) \\ x^{2} - 2x - 8 < 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x^{2} + 4x - 5 = 0\ \ \]

\[D_{1} = 2^{2} + 5 = 9\]

\[x_{1,2} = - 2 \pm 3 = - 5;1.\]

\[(2)\ x^{2} - 2x - 8 = 0\ \ \]

\[D_{1} = 1 + 8 = 9\]

\[x_{1,2} = 1 \pm 3 = - 2;4.\]

\[\left\{ \begin{matrix} (x + 5)(x - 1) > 0 \\ (x + 2)(x - 4) < 0 \\ \end{matrix} \right.\ \]

\[x \in (1;4).\]