\[\begin{array}{l}
{\text{Решить уравнение:}} \hfill \\
\sqrt {{{\left( {x + y} \right)}^2} + {{\left( {2x - y} \right)}^2} + 1} + \sqrt {{{\left( {2y - x} \right)}^2} + {{\left( {y + 1} \right)}^2} + 9} = \sqrt {9{y^2} + {{\left( {2x + 1} \right)}^2} + 16} \hfill \\
\end{array}\]
\[\begin{array}{l}
\overrightarrow a = \left\{ {x + y;2x - y;1} \right\} \hfill \\
\overrightarrow b = \left\{ {2y - x;y + 1;3} \right\} \hfill \\
\overrightarrow a + \overrightarrow b = \left\{ {3y;2x + 1;4} \right\} \hfill \\
{\text{Исходное уравнение имеет вид }}\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| = \left| {\overrightarrow a + \overrightarrow b } \right| \Rightarrow \overrightarrow a \parallel \overrightarrow b \Rightarrow \hfill \\
\frac{{x + y}}{{2y - x}} = \frac{{2x - y}}{{y + 1}} = \frac{1}{3} \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{1}{{22}} \hfill \\
y = - \frac{2}{{11}} \hfill \\
\end{array} \right. \hfill \\
\end{array}\]
\[\left( {\frac{1}{{22}}; - \frac{2}{{11}}} \right)\]
