Đáp án:
\(\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\\
y = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:y \ne - \dfrac{1}{2}\\
\left\{ \begin{array}{l}
\left| {x - 3} \right| + \dfrac{5}{{2y + 1}} = 2\\
2\left| {x - 3} \right| - \dfrac{1}{{2y + 1}} = \dfrac{9}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left| {x - 3} \right| + \dfrac{5}{{2y + 1}} = 2\\
10\left| {x - 3} \right| - \dfrac{5}{{2y + 1}} = 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11\left| {x - 3} \right| = 11\\
2\left| {x - 3} \right| - \dfrac{1}{{2y + 1}} = \dfrac{9}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left| {x - 3} \right| = 1\\
\dfrac{1}{{2y + 1}} = \dfrac{1}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x - 3 = 1\\
x - 3 = - 1
\end{array} \right.\\
2y + 1 = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\\
y = 2
\end{array} \right.
\end{array}\)
