Решебник по алгебре 9 класс Мерзляк Задание 449

Авторы:
Год:2023
Тип:учебник
Серия:Алгоритм успеха

Задание 449

\[\boxed{\text{449\ (449).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} x + y = 5 \\ xy = 6\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 5 - x \\ y = \frac{6}{x}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[y = 5 - x\]

\[x\] \[1\] \[2\]
\[y\] \[4\] \[3\]

\[y = \frac{6}{x}\]

\[x\] \[1\] \[2\] \[3\] \[6\] \[- 1\] \[- 2\] \[- 3\] \[- 6\]
\[y\] \[6\] \[3\] \[2\] \[1\] \[- 6\] \[- 3\] \[- 2\] \[- 1\]

\[Ответ:(2;3);\text{\ \ }(3;2).\]

\[2)\ \left\{ \begin{matrix} y + x^{2} = 3 \\ y = x - 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 - x^{2} \\ y = x - 1\ \ \\ \end{matrix} \right.\ \]

\[y = 3 - x^{2}\]

\[график\ функции\ y = - x^{2}\ \]

\[поднять\ на\ 3\ единицы\ ввепх.\]

\[y = x - 1\]

\[x\] \[0\] \[1\]
\[y\] \[- 1\] \[0\]

\[Ответ:( \approx - 2,5;\ - 3,5);\ \ \]

\[( \approx 1,5;\ \approx 0,6).\]

\[3)\ \left\{ \begin{matrix} x^{2} + y^{2} = 4 \\ x + y = 2\ \ \ \\ \end{matrix}\ \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x^{2} + y^{2} = 4 \\ y = 2 - x\ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + y^{2} = 4\]

\[\text{O\ }(0;0)\]

\[r = 2\]

\[y = 2 - x\]

\[x\] \[0\] \[1\]
\[y\] \[2\] \[1\]

\[\]

\[Ответ:(0;2);\ \ (2;0)\]

\[4)\ \left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ xy = - 12\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ y = - \frac{12}{x}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[x^{2} + y^{2} = 25\]

\[\text{O\ }(0;0)\]

\[r = 5\]

\[y = - \frac{12}{x}\ \]

\[x\] \[2\] \[6\] \[1\] \[12\] \[- 2\] \[- 6\] \[- 1\] \[- 12\]
\[y\] \[- 6\] \[- 2\] \[- 12\] \[- 1\] \[6\] \[2\] \[12\] \[1\]

\[Ответ:( - 4;3);\ ( - 3;4);\ \]

\[(4;\ - 3);\ (3;\ - 4).\]

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