ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 227

\[\boxed{\text{228.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[(x + 1)\left( x^{4} - 6x^{2} + 5 \right) = 0\]

\[1)\ x + 1 = 0,\ \ x_{1} = - 1;\]

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

\[Пусть\ x^{2} = t,\ \ t \geq 0:\]

\[t^{2} - 6t + 5 = 0\]

\[D = 3^{2} - 5 = 4\]

\[t_{1,2} = 3 \pm 2 = 1;5;\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 1 \\ x² = 5 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{2,3} = \pm 1\ \ \ \ \\ x_{4,5} = \pm \sqrt{5.} \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 1;\ \ x = \pm \sqrt{5}.\]

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

\[1)\ x - 1 = 0,\ \ x_{1} = 1;\]

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

\[Пусть\ x^{2} = t,\ \ \ t \geq 0:\]

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

\[D = 1 + 3 = 4\]

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

\[Так\ как\ t \geq 0,\ то\ t = 3 \Longrightarrow x^{2} =\]

\[= 3 \Longrightarrow x_{2,3} = \pm \sqrt{3}.\]

\[Ответ:x = 1;\ \ x = \pm \sqrt{3}.\]