Решебник по алгебре 9 класс Мерзляк Задание 816

\[\boxed{\mathbf{816\ (816).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\sqrt{x^{2} + 1} + \sqrt{y^{2} + 1} =\]

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

\[Пусть\ x^{2} + 1 = a,\ \ \]

\[y^{2} + 1 = b,\ \ a \geq 0,\]

\[\ \ b \geq 0,\ \]

\[тогда\ \ \sqrt{x^{2} + 1} = a,\ \ \]

\[x^{2} + 1 = a^{2},\ \ x^{2} = a^{2} - 1,\]

\[\sqrt{y^{2} + 1} = b,\ \ y^{2} + 1 = b^{2},\]

\[\text{\ \ }y^{2} = b^{2} - 1,\ \ тогда\ \]

\[a^{2} - 1 + b^{2} - 1 = a + b\]

\[a^{2} + b^{2} = a + b \Longrightarrow подберем\]

\[\ решения:\ \ (0;0),\ (1;1),\ \]

\[(1;0),\ (0;1).\]

\[Если\ a = b = 0,\]

\[\text{\ \ }то\ \ \sqrt{x^{2} + 1} = 0,\ \ x^{2} \neq - 1,\]

\[\text{\ \ }\sqrt{y^{2} + 1} = 0,\ \ y^{2} \neq - 1.\]

\[Аналогично:\ \ (1;0)\ и\ \ (0;1).\]

\[Если\ \ a = b = 1,\]

\[\text{\ \ }\sqrt{x^{2} + 1} = 1,\ \ x^{2} + 1 = 1,\]

\[\ \ x² = 0,\ \ x = 0.\]

\[\sqrt{y^{2} + 1} = 1,\ \ y^{2} + 1 = 1,\ \]

\[\ y² = 0,\ \ y = 0.\]

\[Ответ:(0;0).\]

\[\boxed{\mathbf{816.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\left\{ \begin{matrix} 3x - y - 5 = 0\ \ \ \\ y = \frac{2}{3}x^{2} - 2x + 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} y = 3x - 5\ \ \ \ \ \ \ \ \ \ \ \\ y = \frac{2}{3}x^{2} - 2x + 4 \\ \end{matrix} \right.\ \]

\[3x - 5 = \frac{2}{3}x^{2} - 2x + 4\ \ \ \ | \cdot 3\]

\[9x - 15 = 2x^{2} - 6x + 12\]

\[2x² - 15x + 27 = 0\]

\[D = 225 - 8 \cdot 27 = 9\]

\[x = \frac{15 + 3}{4} = 4,5\]

\[x = \frac{15 - 3}{4} = 3\]

\[\left\{ \begin{matrix} x = 3 \\ y = 9 - 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 3 \\ y = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \]

\[\ \left\{ \begin{matrix} x = 4,5 \\ y = 8,5 \\ \end{matrix} \right.\ \]

\[Ответ:(3;4);\ \ (4,5;8,5).\]

\[2)\ \left\{ \begin{matrix} 2x - 3y - 3 = 0 \\ xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = \frac{3y + 3}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{(3y + 3)y}{2} = 3\ \ \ | \cdot 2 \\ \end{matrix} \right.\ \]

\[(3y + 3)y = 6\]

\[3y² + 3y = 6\ \ \ |\ :3\]

\[y^{2} + y - 2 = 0\]

\[y = - 2,\ \ y = 1\]

\[\left\{ \begin{matrix} x = 3 \\ y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \left\{ \begin{matrix} x = - 1,5 \\ y = - 2\ \ \\ \end{matrix} \right.\ \]

\[Ответ:(3;1),\ ( - 1,5;\ - 2).\]

\[3)\ \left\{ \begin{matrix} x^{2} + y^{2} = 13 \\ xy = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \left( \frac{6}{y} \right)^{2} + y^{2} = 13 \\ x = \frac{6}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} \frac{36}{y^{2}} + y^{2} - 13 = 0 \\ x = \frac{6}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[y^{4} - 13y^{2} + 36 = 0,\]

\[\text{\ \ }y^{2} = 4,\ \ y^{2} = 9\]

\[\left\{ \begin{matrix} y^{2} = 4 \\ x = \frac{6}{y} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 2 \\ x = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = - 2 \\ x = - 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y^{2} = 9 \\ x = \frac{6}{y} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 \\ x = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

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

\[Ответ:(3;2),\ ( - 3;\ - 2),\ \]

\[(2;3),\ ( - 2;\ - 3).\]