Решебник по алгебре 7 класс Мерзляк Задание 1067

\[\boxed{\text{1067\ (1067).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[x + y = 0\ \ \ и\ \ \ x - 3 = 0\]

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

\[Ответ:(3;\ - 3).\]

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

\[Ответ:(1,5;0,75).\]

\[Ответ:\left( 4;\ - \frac{2}{3} \right).\]

\[4)\ x^{2} + y^{2} + 10x - 12y + 61 =\]

\[= 0\]

\[(x + 5)^{2} + (y - 6)^{2} = 0\]

\[\left\{ \begin{matrix} x + 5 = 0 \\ y - 6 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 5 \\ y = 6\ \\ \end{matrix} \right.\ \]

\[Ответ:( - 5;6).\]

\[(5x - 3y)^{2} + (y + 4)^{2} = 0\]

\[Ответ:( - 2,4;\ - 4)\text{.\ }\]