ГДЗ по алгебре 9 класс Макарычев Задание 283

\[\boxed{\text{283\ (283).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{5} + x^{4} - 6x^{3} - 6x^{2} +\]

\[+ 5x + 5 = 0\]

\[x^{4} \cdot (x + 1) - 6x^{2}(x + 1) +\]

\[+ 5 \cdot (x + 1) = 0\]

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

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

\[x_{1} = - 1;\]

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

\[Пусть\ x^{2} = t;\ \ t \geq 0:\]

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

\[D_{1} = 3^{2} - 5 = 4\]

\[t_{1} = 3 + 2 = 5;\ \ \ t_{2} = 3 - 2 = 1.\]

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

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

\[\textbf{б)}\ x^{5} - x^{4} - 2x^{3} + 2x^{2} -\]

\[- 3x + 3 = 0\]

\[x^{4} \cdot (x - 1) - 2x^{2}(x - 1) -\]

\[- 3 \cdot (x - 1) = 0\]

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

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

\[x_{1} = 1.\]

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

\[Пусть\ x^{2} = t;\ \ \ t \geq 0:\]

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

\[D_{1} = 1 + 3 = 4\]

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

\[Так\ как\ t \geq 0,\ то\ t = 3:\]

\[x^{2} = 3\]

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

\[Ответ:x = 1;\ \ x = \pm \sqrt{3}.\]