ГДЗ по алгебре 9 класс Макарычев Задание 282

\[\boxed{\text{282\ (282).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( x^{2} - 1 \right)\left( x^{2} + 1 \right) -\]

\[- 4 \cdot \left( x^{2} - 11 \right) = 0\]

\[x^{4} - 1 - 4x^{2} + 44 = 0\]

\[x^{4} - 4x^{2} + 43 = 0\]

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

\[t^{2} - 4t + 43 = 0\]

\[D_{1} = 4 - 43 = - 39 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:нет\ корней.\]

\[\textbf{б)}\ 3x^{2}(x - 1)(x + 1) - 10x^{2} +\]

\[+ 4 = 0\]

\[3x^{2}\left( x^{2} - 1 \right) - 10x^{2} + 4 = 0\]

\[3x^{4} - 3x^{2} - 10x^{2} + 4 = 0\]

\[3x^{4} - 13x^{2} + 4 = 0\]

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

\[3t^{2} - 13t + 4 = 0\]

\[D = 13^{2} - 4 \cdot 3 \cdot 4 =\]

\[= 169 - 48 = 121\]

\[t_{1} = \frac{13 + 11}{6} = 4;\ \]

\[\ t_{2} = \frac{13 - 11}{6} = \frac{1}{3}.\]

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

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