Решебник по алгебре 9 класс Мерзляк Задание 141

 

\[\boxed{\text{141\ (141).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\frac{|y + 7|}{y + 7} = 1;\ \ y \neq - 7\]

\[при\ y + 7 \geq :\]

\[|y + 7| = y + 7\]

\[\frac{y + 7}{y + 7} = 1\]

\[y + 7 > 0\]

\[y > - 7.\]

\[Ответ:x \in ( - 7;\ + \infty).\]

\[2)\ \frac{|6 - y|}{y - 6} = 1;\ \ \ y \neq 6\]

\[При\ 6 - y < 0:\]

\[|6 - y| = - (6 - y) = y - 6.\]

\[\frac{y - 6}{y - 6} = 1\]

\[6 - y < 0\]

\[y > 6.\]

\[Ответ:x \in (6; + \infty).\]

\[\boxed{\mathbf{141.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[= 0\]

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

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

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

\[Ответ:\ x = - 1;\ x = - 5.\]

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

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

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

\[Ответ:x = 1;\ x = - 1;x = 0;\ \]

\[x = - 2.\]

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

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

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

\[Ответ:нет\ корней.\]

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

\[= 0\]

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

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

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

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