ГДЗ по алгебре 9 класс Макарычев Задание 323

 

\[\boxed{\text{323\ (323).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y^{4} - 24y^{2} - 25 = 0\]

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

\[t^{2} - 24t - 25 = 0\]

\[D_{1} = 144 + 25 = 169\]

\[t_{1,2} = 12 \pm 13 = 25; - 1.\]

\[Так\ как\ t > 0,\ то\ t = 25:\]

\[y^{2} = 25\]

\[y = \pm 5.\]

\[Ответ:y = \pm 5.\]

\[\textbf{б)}\ x^{4} - 9x^{2} + 18 = 0\]

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

\[t^{2} - 9t + 18 = 0\]

\[D = 81 - 4 \cdot 18 = 9\]

\[t_{1,2} = \frac{9 \pm 3}{2} = 3;6.\]

\[\left\{ \begin{matrix} x^{2} = 3 \\ x^{2} = 6 \\ \end{matrix} \Longrightarrow \right.\ \left\{ \begin{matrix} x_{1,2} = \pm \sqrt{3} \\ x_{3,4} = \pm \sqrt{6}. \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm \sqrt{3};\ \ x = \pm \sqrt{6}.\]

\[\boxed{\text{323.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 2x^{7} + x^{6} + 2x^{4} +\]

\[+ x^{3} + 2x + 1 = 0\]

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

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

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

\[1)\ 2x + 1 = 0\ \ \]

\[x_{1} = - \frac{1}{2} = - 0,5.\]

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

\[Пусть\ t = x^{3};\ t^{2} = x^{6}:\ \ \]

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

\[D = 1 - 4 < 0 \Longrightarrow корней\ нет.\]

\[Ответ:x = - 0,5.\]

\[\textbf{б)}\ x^{7} - 2x^{6} + 2x^{4} - 4x^{3} +\]

\[+ x - 2 = 0\]

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

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

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

\[1)\ x - 2 = 0\ \ \]

\[x_{1} = 2.\]

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

\[Пусть\ \ t = x^{3};\ \ t^{2} = x^{6}:\]

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

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

\[t = - 1\]

\[\Longrightarrow x^{3} = - 1\ \ \]

\[x_{2} = - 1.\]

\[Ответ: - 1;2.\]