ГДЗ по алгебре 9 класс Макарычев Задание 278

\[\boxed{\text{278\ (278).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{4} - 5x^{2} - 36 = 0\]

\[Пусть\ \ x^{2} = t;\ \ \ x^{4} = t^{2};\ \ t \geq 0:\]

\[t^{2} - 5t - 36 = 0\]

\[t_{1} + t_{2} = 5;\ \ \ t_{1} \cdot t_{2} = - 36\]

\[t_{1} = 9;\ \ \ \ \ t_{2} = - 4 \Longrightarrow не\]

\[\ подходит.\]

\[x^{2} = 9\ \]

\[x = \pm 3.\]

\[Ответ:x = \pm 3.\]

\[\textbf{б)}\ y^{4} - 6y^{2} + 8 = 0\]

\[Пусть\ \ y^{2} = t;\ \ y^{4} = t^{2};\ \]

\[\ t \geq 0.\]

\[t^{2} - 6t + 8 = 0\]

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

\[t_{1} = 3 + 1 = 4;\ \ \ \ \ \]

\[t_{2} = 3 - 1 = 2;\]

\[\left\{ \begin{matrix} y^{2} = 4 \\ y^{2} = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{1,2} = \pm 2 \\ y_{3,4} = \pm \sqrt{2} \\ \end{matrix} \right.\ .\]

\[Ответ:y = \pm 2;\ \ y = \pm \sqrt{2}.\]

\[\textbf{в)}\ t^{4} + 10t^{2} + 25 = 0\]

\[Пусть\ t^{2} = a;\ \ t^{4} = a^{2};\ \ a \geq 0.\]

\[a^{2} + 10a + 25 = 0\]

\[(a + 5)^{2} = 0\]

\[a + 5 = 0\]

\[a = - 5 \Longrightarrow не\ подходит.\]

\[Ответ:корней\ исходное\ \]

\[уравнение\ не\ имеет.\]

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

\[Пусть\ x^{2} = a;\ \ x^{4} = a^{2};\ \ a \geq 0.\]

\[4a^{2} - 5a + 1 = 0\]

\[D = 25 - 16 = 9\]

\[a_{1} = \frac{5 + 3}{8} = 1;\ \ \ a_{2} = \frac{5 - 3}{8} = \frac{1}{4}.\]

\[\left\{ \begin{matrix} x^{2} = 1 \\ x^{2} = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 1 \\ x_{3,4} = \pm \frac{1}{2} \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 1;\ \ x = \pm \frac{1}{2}.\]

\[\textbf{д)}\ 9x^{4} - 9x^{2} + 2 = 0\]

\[Пусть\ \ x^{2} = t;\ \ t \geq 0.\]

\[9t^{2} - 9t + 2 = 0\]

\[D = 81 - 72 = 9\]

\[t_{1} = \frac{9 - 3}{18} = \frac{1}{3};\ \ t_{2} = \frac{9 + 3}{18} = \frac{2}{3}.\]

\[x^{2} = \frac{1}{3} \Longrightarrow x = \pm \sqrt{\frac{1}{3}}.\]

\[x^{2} = \frac{2}{3} \Longrightarrow x = \pm \sqrt{\frac{2}{3}}.\]

\[Ответ:\ x = \pm \sqrt{\frac{2}{3}};\ \ x = \pm \sqrt{\frac{1}{3}}.\]

\[\textbf{е)}\ 16y^{4} - 8y^{2} + 1 = 0\]

\[Пусть\ \ y^{2} = t;\ \ \ t \geq 0.\]

\[16t^{2} - 8t + 1 = 0\]

\[(4t - 1)^{2} = 0\]

\[4t - 1 = 0\]

\[4t = 1\]

\[t = \frac{1}{4}.\]

\[y^{2} = \frac{1}{4}\]

\[y = \pm \frac{1}{2}.\]

\[Ответ:y = \pm \frac{1}{2}.\]