Đáp án:
\(y = \dfrac{{11 - \sqrt {21} }}{4};x = \dfrac{{13 + \sqrt {21} }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {x - 1} - \sqrt {2y} = 1\\
x = 12 - 2y
\end{array} \right.\left( {DK:x \ge 1;y \ge 0} \right)\\
\to \left\{ \begin{array}{l}
x = 12 - 2y\\
\sqrt {12 - 2y - 1} = \sqrt {2y} + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 12 - 2y\\
\sqrt {11 - 2y} = 1 + \sqrt {2y}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 12 - 2y\\
11 - 2y = 1 + 2\sqrt {2y} + 2y\left( {DK:0 \le y \le \dfrac{{11}}{2}} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 12 - 2y\\
4y + 2\sqrt {2y} - 10 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt y = \dfrac{{\sqrt {42} - \sqrt 2 }}{4}\\
\sqrt y = \dfrac{{ - \sqrt {42} - \sqrt 2 }}{4}\left( l \right)
\end{array} \right.\\
\to y = \dfrac{{11 - \sqrt {21} }}{4}\left( {TM} \right)\\
\to x = \dfrac{{13 + \sqrt {21} }}{2}\left( {TM} \right)
\end{array}\)
