Решебник по алгебре 9 класс Макарычев Задание 395

 

\[\boxed{\text{395\ (395).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[( - 1;3)\]

\[\textbf{а)}\ x^{2} - y + 2 = 0\]

\[( - 1)^{2} - 3 + 2 = 0\ \ \]

\[0 = 0 \Longrightarrow является;\]

\[\textbf{б)}\ xy + y = 6\]

\[- 1 \cdot 3 + 3 = 6\ \ \]

\[0 \neq 6 \Longrightarrow не\ является;\]

\[\textbf{в)}\ x^{2} + y^{2} = 10\]

\[( - 1)^{2} + 3^{2} = 10\]

\[10 = 10 \Longrightarrow является;\]

\[\textbf{г)}\ x^{2} - y^{2} + 8 = 0\]

\[( - 1)^{2} - 3^{2} + 8 = 0\]

\[0 = 0 \Longrightarrow является.\]

\[\boxed{\text{395.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\text{M\ }(1;2);\ \ y = kx:\]

\[2 = k \cdot 1\]

\[k = 2\]

\[y = 2x.\]

\[\left\{ \begin{matrix} (x - 4)^{2} + (y - 6)^{2} = 25 \\ y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 4)^{2} + (2x - 6)^{2} = 25 \\ y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 8x + 16 + 4x^{2} - 24x + 36 = 25 \\ y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} 5x^{2} - 32x + 27 = 0 \\ y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[5x^{2} - 32x + 27 = 0\]

\[D_{1} = 16^{2} - 27 \cdot 5 =\]

\[= 256 - 135 = 121\]

\[x_{1,2} = \frac{16 \pm 11}{5} = 1;5,4.\]

\[\left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} x_{2} = 5,4\ \ \ \\ y_{2} = 10,8. \\ \end{matrix} \right.\ \]

\[Ответ:точка\ (5,4;10,8).\]