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

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

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

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

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

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

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

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

\[Ответ:( - 0,4; - 0,2).\]