Đáp án:
\(y = - \dfrac{1}{3};x = \dfrac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge \dfrac{1}{2};y \ge - \dfrac{3}{2}\\
\left\{ \begin{array}{l}
\sqrt {2x - 1} + \sqrt {3y + 2} = 3\\
x = \dfrac{{7 + 6y}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt {2.\dfrac{{7 + 6y}}{2} - 1} + \sqrt {3y + 2} = 3\\
x = \dfrac{{7 + 6y}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt {7 + 6y - 1} + \sqrt {3y + 2} = 3\\
x = \dfrac{{7 + 6y}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt {6 + 6y} + \sqrt {3y + 2} = 3\left( 1 \right)\\
x = \dfrac{{7 + 6y}}{2}
\end{array} \right.\\
\left( 1 \right) \to 6 + 6y + 2\sqrt {\left( {6 + 6y} \right)\left( {3y + 2} \right)} + 3y + 2 = 9\\
\to 2\sqrt {18{y^2} + 30y + 12} = 1 - 9y\\
\to 4\left( {18{y^2} + 30y + 12} \right) = 1 - 18y + 81{y^2}\left( {DK:\dfrac{1}{9} \ge y} \right)\\
\to 9{y^2} - 138y - 47 = 0\\
\to \left[ \begin{array}{l}
y = \dfrac{{47}}{3}\left( l \right)\\
y = - \dfrac{1}{3}\left( {TM} \right)
\end{array} \right.\\
\to x = \dfrac{5}{2}
\end{array}\)
