Решите уравнение: (3x-y+1)^2+x^2-4xy+4y^2=0.

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

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

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

\[\left\{ \begin{matrix} 3x + 1 = y\ \ \ \ \ \ \ \\ x - 6x - 2 = 0 \\ \end{matrix} \right.\ \]

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

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

\[\left\{ \begin{matrix} y = - \frac{1}{5} \\ x = - \frac{2}{5} \\ \end{matrix} \right.\ \]

\[Ответ:( - 0,4; - 0,2).\]