Đáp án:
1) \(\left\{ \begin{array}{l}
x = 9\\
\left[ \begin{array}{l}
y = \dfrac{5}{8}\\
y = \dfrac{3}{8}
\end{array} \right.
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0;y \ne \dfrac{1}{2}\\
\left\{ \begin{array}{l}
\dfrac{8}{{\sqrt x + 1}} + \dfrac{3}{{\left| {2y - 1} \right|}} = 5\\
\dfrac{4}{{\sqrt x + 1}} + \dfrac{1}{{\left| {2y - 1} \right|}} = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{8}{{\sqrt x + 1}} + \dfrac{3}{{\left| {2y - 1} \right|}} = 5\\
\dfrac{{12}}{{\sqrt x + 1}} + \dfrac{3}{{\left| {2y - 1} \right|}} = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{4}{{\sqrt x + 1}} = 1\\
\dfrac{4}{{\sqrt x + 1}} + \dfrac{1}{{\left| {2y - 1} \right|}} = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt x + 1 = 4\\
\dfrac{1}{{\left| {2y - 1} \right|}} = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 9\\
\left| {2y - 1} \right| = \dfrac{1}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 9\\
\left[ \begin{array}{l}
2y - 1 = \dfrac{1}{4}\\
2y - 1 = - \dfrac{1}{4}
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 9\\
\left[ \begin{array}{l}
y = \dfrac{5}{8}\\
y = \dfrac{3}{8}
\end{array} \right.
\end{array} \right.
\end{array}\)
