Решебник по алгебре 8 класс Мерзляк ФГОС Задание 779

 

\[\boxed{\mathbf{779\ (779).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (x + 3)^{4} - 3 \cdot (x + 3)^{2} - 4 =\]

\[= 0\]

\[\left\{ \begin{matrix} (x + 3)^{2} = t\ \ \ \ \ \ \\ t^{2} - 3t - 4 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 3,\ \ t_{1} \cdot t_{2} = - 4,\ \ \]

\[t_{1} = 4,\ \ t_{2} = - 1\]

\[Ответ:\ x = - 1;\ x = - 5.\]

\[\left\{ \begin{matrix} (2x + 1)^{2} = t\ \ \ \ \\ t² - 10t + 9 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 10,\ \ t_{1} \cdot t_{2} = 9,\ \ \]

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

\[Ответ:x = 1;\ x = - 1;x = 0;\ \]

\[x = - 2.\]

\[\left\{ \begin{matrix} (6x - 7)^{2} = t\ \ \\ t^{2} + 4t + 3 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = - 4,\ \ t_{1} \cdot t_{2} = 3,\ \ \]

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

\[Ответ:нет\ корней.\]

\[4)\ (x - 4)^{4} + 2 \cdot (x - 4)^{2} - 8 =\]

\[= 0\]

\[\left\{ \begin{matrix} (x - 4)^{2} = t\ \ \ \ \ \\ t^{2} + 2t - 8 = 0 \\ \end{matrix} \right.\ \]

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

\[t_{1} = - 4,\ \ t_{2} = 2\]

\[Ответ:x = 4 + \sqrt{2};\ \ \ x = 4 - \sqrt{2}.\]

\[\boxed{\mathbf{7}\mathbf{7}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[x^{2} + bx + c = 0\]

\[x_{1} \cdot x_{2} = c\]

\[x_{1} + x_{2} = - b\]

\[Если\ x_{1} = b;\ \ x_{2} = c:\]

\[bc = c,\ \ b = 1\]

\[b + c = - b \Longrightarrow c = - 2\]

\[или\ \]

\[b = c = 0\]

\[Ответ:b = 1;\ \ c = - 2\ \ \ \ или\ \ \ \]

\[b = 0;\ \ c = 0.\]