Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;y \ne 1\\
\left\{ \begin{array}{l}
2\sqrt x - \dfrac{1}{{\left| {y - 1} \right|}} = 4\\
\sqrt x - \dfrac{1}{{\left| {1 - y} \right|}} = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2\sqrt x - \dfrac{1}{{\left| {y - 1} \right|}} = 4\\
\sqrt x - \dfrac{1}{{\left| {y - 1} \right|}} = - 2\left( {do:\left| {y - 1} \right| = \left| {1 - y} \right|} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2\sqrt x - \sqrt x = 4 - \left( { - 2} \right)\\
\dfrac{1}{{\left| {y - 1} \right|}} = \sqrt x + 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt x = 6\\
\dfrac{1}{{\left| {y - 1} \right|}} = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 36\left( {tmdk} \right)\\
\left| {y - 1} \right| = \dfrac{1}{8}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 36\\
\left[ \begin{array}{l}
y - 1 = \dfrac{1}{8}\\
y - 1 = - \dfrac{1}{8}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 36\\
\left[ \begin{array}{l}
y = \dfrac{9}{8}\\
y = \dfrac{7}{8}
\end{array} \right.
\end{array} \right.\\
Vay\,\left( {x;y} \right) = \left\{ {\left( {36;\dfrac{9}{8}} \right);\left( {36;\dfrac{7}{8}} \right)} \right\}
\end{array}$
