Решебник по алгебре 9 класс Макарычев Задание 218

 

\[\boxed{\text{218}\text{\ (218)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} + px + q;\ \ \ x_{1} = p;\ \ \ x_{2} = q.\]

\[x^{2} + px + q = (x - p)(x - q)\]

\[(x - p)(x - q) = x^{2} - qx - px +\]

\[+ qp = x^{2} - (p + q)x + pq.\]

\[По\ теореме\ Виета:\]

\[\left\{ \begin{matrix} pq = q\ \ \ \ \ \ \ \ \ \ \ \ \\ - (p + q) = p \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} p = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ - p - q = p \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} p = 1\ \ \ \\ q = - 2 \\ \end{matrix} \right.\ \]

\[\Longrightarrow x^{2} + x - 2 \Longrightarrow искомый\ \]

\[трехчлен.\]

\[\boxed{\text{218.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \ 3x³ - x^{2} + 18x - 6 = 0\]

\[x^{2}(3x - 1) + 6 \cdot (3x - 1) = 0\]

\[\left( x^{2} + 6 \right)(3x - 1) = 0\]

\[x_{1}^{2} = - 6 \Longrightarrow корней\ нет;\]

\[3x_{2} = 1\ \ \]

\[x_{2} = \frac{1}{3}\]

\[Ответ:x = \frac{1}{3}.\]

\[\textbf{б)}\ 2x^{4} - 18x^{2} = 5x^{3} - 45x\]

\[2x^{2}\left( x^{2} - 9 \right) = 5x\left( x^{2} - 9 \right)\]

\[\left( 2x^{2} - 5x \right)\left( x^{2} - 9 \right) = 0\]

\[x(2x - 5)(x - 3)(x + 3) = 0\]

\[x_{1} = 0,\ \ x_{2} = 2,5;\ \]

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

\[Ответ:x = 0;\ \ x = 2,5;\ \ x = \pm 3.\]