ГДЗ по алгебре 9 класс Макарычев Задание 537

 

\[\boxed{\text{537\ (537).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[По\ условию\ коэффициенты\ при\ \]

\[x^{2}\ и\ \text{x\ }\ равны\ 8\ и\ ( - 2):\]

\[\left\{ \begin{matrix} b - 3a = 8\ \ \\ 2 + ab = - 2 \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} b = 3a + 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a(3a + 8) + 4 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} b = 3a + 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3a² + 8a + 4 = 0 \\ \end{matrix} \right.\ \ \]

\[3a^{2} + 8a + 4 = 0\]

\[D = 16 - 12 = 4\]

\[a_{1,2} = \frac{- 4 \pm 2}{3};\]

\[\left\{ \begin{matrix} a_{1} = - 2 \\ b_{1} = 2\ \ \ \\ \end{matrix} \right.\ \ \ или\ \ \ \left\{ \begin{matrix} a_{2} = - \frac{2}{3} \\ b_{2} = 6\ \ \ \ . \\ \end{matrix} \right.\ \]

\[Ответ:a = - 2,\ b = 2\ \ или\]

\[a = - \frac{2}{3},b = 6.\]

\[\boxed{\text{537.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 4x^{4} + 4x² - 15 = 0\]

\[Пусть\ t = x^{2},\ t \geq 0,\ тогда:\]

\[4t^{2} + 4t - 15 = 0\]

\[D = 4 + 4 \cdot 15 = 64\]

\[t_{1,2} = \frac{- 2 \pm 8}{4},\ \ так\ как\]

\[\ t \geq 0,\ то\ t = 1,5.\]

\[\Longrightarrow x^{2} = 1,5 \Longrightarrow x = \pm \sqrt{1,5}.\]

\[Ответ:x = \pm \sqrt{1,5}.\]

\[\textbf{б)}\ 2x^{4} - x^{2} - 36 = 0\]

\[Пусть\ t = x^{2},\ t \geq 0,\ тогда:\]

\[2t^{2} - t - 36 = 0\]

\[D = 1 + 4 \cdot 2 \cdot 36 = 289\]

\[t_{1,2} = \frac{1 \pm 17}{4},\ \ так\ как\ \]

\[t \geq 0,\ то\ \ t = 4,5;\]

\[x² = 4,5 \Longrightarrow x = \pm \sqrt{4,5}.\]

\[Ответ:x = \pm \sqrt{4,5}.\]