Решебник по алгебре 9 класс Макарычев Задание 447

 

\[\boxed{\text{447\ (447).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 12 \\ xy = - 6\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - \frac{6}{y}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ \frac{36}{y^{2}} + y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - \frac{6}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 36 + y^{4} - 12y^{2} = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - \frac{6}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ \left( y^{2} - 6 \right)^{2} = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = \sqrt{6}\text{\ \ \ \ } \\ x_{1} = - \sqrt{6} \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \]

\[\left\{ \begin{matrix} y_{2} = - \sqrt{6} \\ x_{2} = \sqrt{6.}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - y^{2} = 34 \\ xy = 20\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{20}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 2 \cdot \frac{400}{y^{2}} - y^{2} = 34 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{20}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 800 - y^{4} - 34y^{2} = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{20}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y^{4} + 34y^{2} - 800 = 0 \\ \end{matrix} \right.\ \]

\[Пусть\ t = y^{2};\ \ t \geq 0:\]

\[t^{2} + 34t - 800 = 0\]

\[D = 17^{2} + 800 = 1089\]

\[t_{1,2} = - 17 \pm 33\]

\[так\ как\ t \geq 0:\]

\[t = 16 \Longrightarrow y^{2} = 16.\]

\[1)\ y_{1} = 4;\ \ x_{1} = 5;\]

\[2)y_{2} = - 4;\ \ x_{2} = - 5.\]

\[Ответ:а)\ \left( - \sqrt{6};\sqrt{6} \right);\ \ \left( \sqrt{6};\ - \sqrt{6} \right);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (5;4);\ \ ( - 5;\ - 4).\]

\(\boxed{\text{447.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

\[\textbf{а)}\ 2x - 3y + 16 > 0\]

\[2 \cdot ( - 2) - 3 \cdot 3 + 16 > 0\]

\[- 4 - 9 + 16 > 0\]

\[3 > 0 \Longrightarrow да.\]

\[\textbf{б)}\ x^{2} + 3xy - y^{2} < 20\]

\[( - 2)^{2} + 3 \cdot ( - 2) \cdot 3 - 3^{2} < 20\]

\[4 - 18 - 9 < 20\]

\[- 23 < 20 \Longrightarrow да.\]

\[\textbf{в)}\ (x + 3)^{2} + (y - 4)^{2} < 2\]

\[( - 2 + 3)^{2} + (3 - 4)^{2} < 2\]

\[1 + 1 < 2\]

\[2 < 2 \Longrightarrow нет.\]

\[\textbf{г)}\ (x + y)(y - 8) < 1\]

\[( - 2 + 3)(3 - 8) < 1\]

\[1 \cdot ( - 5) < 1\]

\[- 5 < 1 \Longrightarrow да.\]

\[\textbf{д)}\ x^{2} + y^{2} - x - y \geq 0\]

\[( - 2)^{2} + 3^{2} + 2 - 3 \geq 0\]

\[4 + 9 - 1 \geq 0\]

\[12 \geq 0 \Longrightarrow да.\]

\[\textbf{е)}\ 3x^{2} - 5y^{2} + x - y < 11\]

\[3 \cdot ( - 2)^{2} - 5 \cdot 3^{2} - 2 - 3 < 11\]

\[12 - 45 - 5 < 11\]

\[- 38 < 11 \Longrightarrow да.\ \]