Решебник по алгебре 9 класс Мерзляк Задание 392

\[\boxed{\text{392\ (392).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ x^{4} - 13x^{2} + 36 = 0\]

\[\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - 13t + 36 = 0\ \ \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 13,\ \ t_{1} \cdot t_{2} = 36,\ \ \]

\[t_{1} = 9,\ \ t_{2} = 4.\]

\[\left\{ \begin{matrix} x^{2} = t \\ t = 9 \\ t = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} = 9 \\ x^{2} = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \pm 3 \\ x = \pm 2 \\ \end{matrix} \right.\ \]

\[Ответ:\ x = \pm 3;\ x = \pm 2.\]

\[2)\ x^{4} - 5x^{2} - 6 = 0\]

\[\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - 5t - 6 = 0\ \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 5,\ \ t_{1} \cdot t_{2} = - 6,\ \ \]

\[t_{1} = 6,\ \ t_{2} = - 1\]

\[\left\{ \begin{matrix} x^{2} = t \\ t = 6\ \ \ \\ t = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} = 6\ \ \ \\ x^{2} \neq - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ x = \pm \sqrt{6}\]

\[Ответ:\ x = \pm \sqrt{6}.\]

\[3)\ x^{4} + 9x^{2} + 8 = 0\]

\[\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} + 9t + 8 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = - 9,\ \ t_{1} \cdot t_{2} = 8,\ \ \]

\[t_{1} = - 8,\ \ t_{2} = - 1\]

\[\left\{ \begin{matrix} x^{2} = - 8 \\ x^{2} = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[4)\ x^{4} - 16x^{2} = 0\]

\[x^{2}\left( x^{2} - 16 \right) = 0\]

\[x^{2} = 0\ \ \ \ \ \ \ \ \ \ \ \ x^{2} = 16\]

\[x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \pm 4.\]

\[Ответ:x = 0;x = \pm 4.\]