Đáp án:
\(\left\{ \begin{array}{l}
x = \frac{8}{3}\\
y = \frac{4}{9}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {x - 2y} = 3y\left( 1 \right)\\
\sqrt {x + \sqrt {x - 2y} } = x + \sqrt {x - 2y} - 2\left( * \right)
\end{array} \right.\\
Đặt:\sqrt {x + \sqrt {x - 2y} } = t\left( {t > 0} \right)\\
\to x + \sqrt {x - 2y} = {t^2}\\
\left( * \right) \to t = {t^2} - 2\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 1\left( l \right)
\end{array} \right.\\
\to \sqrt {x + \sqrt {x - 2y} } = 2\\
\to x + \sqrt {x - 2y} = 4\\
\to x + 3y = 4\\
\to x = 4 - 3y\\
\left( 1 \right) \to \sqrt {4 - 3y - 2y} = 3y\\
\to \left\{ \begin{array}{l}
y \ge 0\\
4 - 3y - 2y = 9{y^2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y \ge 0\\
\left[ \begin{array}{l}
y = \frac{4}{9}\\
y = - 1\left( l \right)
\end{array} \right.
\end{array} \right.\\
\to x = \frac{8}{3}
\end{array}\)
