Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
\frac{3}{{\sqrt {y - 2} }} + x - 2y = 5\\
\frac{1}{{\sqrt {y - 2} }} - 2.\left( {x - 2y} \right) = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{3}{{\sqrt {y - 2} }} + x - 2y = 5\\
\frac{3}{{\sqrt {y - 2} }} - 6.\left( {x - 2y} \right) = 12
\end{array} \right.\\
\Rightarrow 7.\left( {x - 2y} \right) = - 7\\
\Rightarrow x - 2y = - 1\\
\Rightarrow x = 2y - 1\\
\Rightarrow \frac{1}{{\sqrt {y - 2} }} - 2.\left( { - 1} \right) = 4\\
\Rightarrow \frac{1}{{\sqrt {y - 2} }} = 2\\
\Rightarrow \sqrt {y - 2} = \frac{1}{2}\\
\Rightarrow y - 2 = \frac{1}{4}\\
\Rightarrow y = \frac{9}{4}\\
\Rightarrow x = 2y - 1 = \frac{9}{2} - 1 = \frac{7}{2}\\
Vậy\,\left( {x;y} \right) = \left( {\frac{7}{2};\frac{9}{4}} \right)
\end{array}$
