ГДЗ по алгебре 9 класс Макарычев Задание 444

\[\boxed{\text{444\ (444).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} y = x^{2} - 8x + 16 \\ 2x - 3y = 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y = x^{2} - 8x + 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - 3 \cdot \left( x^{2} - 8x + 16 \right) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x^{2} - 8x + 16\ \ \ \ \ \ \ \\ - 3x^{2} + 26x - 48 = 0 \\ \end{matrix} \right.\ \]

\[3x^{2} - 26x + 48 = 0\]

\[D = 13^{2} - 3 \cdot 48 = 25\]

\[x_{1.2} = \frac{13 \pm 5}{3} = 6;2\frac{2}{3};\]

\[1)\ x_{1} = 6;\ \ y_{1} = 4;\]

\[2)\ x_{2} = 2\frac{2}{3};\ \ \ y_{2} = 1\frac{7}{9}.\]

\[Ответ:пересекает\ в\ точках\ \]

\[(6;4);\left( 2\frac{2}{3};1\frac{7}{9} \right).\]

\[\textbf{б)}\left\{ \begin{matrix} (x - 5)^{2} + (y - 4)^{2} = 65 \\ 3x - y + 6 = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 3x + 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 5)^{2} + (3x + 6 - 4)^{2} = 65 \\ \end{matrix} \right.\ \ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 3x + 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 10x^{2} + 2x - 36 = 0 \\ \end{matrix} \right.\ \]

\[5x^{2} + x - 18 = 0\]

\[D = 1 + 4 \cdot 5 \cdot 18 = 361\]

\[x_{1,2} = \frac{- 1 \pm 19}{10} = - 2;1,8.\]

\[1)\ x_{1} = - 2;\ \ y_{1} = 0;\]

\[2)\ x_{2} = 1,8;\ \ \ \ y_{2} = 11,4\text{.\ }\]

\[Ответ:пересекаются\ в\ \]

\[точках\ ( - 2;0);(1,8;11,4).\]