Đáp án:
$\begin{array}{l}
Dkxd:\dfrac{{x + 1}}{{2y - 1}} > 0\\
Đặt:\sqrt {\dfrac{{x + 1}}{{2y - 1}}} = t\left( {t > 0} \right)\\
\Rightarrow \sqrt {\dfrac{{2y - 1}}{{x + 1}}} = \dfrac{1}{t}\\
\Rightarrow t + \dfrac{1}{t} = 2\\
\Rightarrow {t^2} - 2t + 1 = 0\\
\Rightarrow {\left( {t - 1} \right)^2} = 0\\
\Rightarrow t = 1\left( {tmdk} \right)\\
\Rightarrow \sqrt {\dfrac{{x + 1}}{{2y - 1}}} = 1\\
\Rightarrow \dfrac{{x + 1}}{{2y - 1}} = 1\\
\Rightarrow x + 1 = 2y - 1\\
\Rightarrow x - 2y = - 2\\
\Rightarrow \left\{ \begin{array}{l}
x - 2y = - 2\\
2x + y = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x - 4y = - 4\\
2x + y = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5y = 3\\
x - 2y = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = \dfrac{3}{5}\\
x = 2y - 2 = \dfrac{{ - 4}}{5}
\end{array} \right.\\
Vay\,\left( {x;y} \right) = \left( { - \dfrac{4}{5};\dfrac{3}{5}} \right)
\end{array}$
