Đáp án:
$\begin{array}{l}
Dkxd:x \ge 2;y \ge - 3\\
\left\{ \begin{array}{l}
\sqrt {x - 2} + \sqrt {y + 3} = 4\\
x + y = 7
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 2} + \sqrt {y + 3} = 4\\
x - 2 + y + 3 = 8
\end{array} \right.\\
Dat:\left\{ \begin{array}{l}
\sqrt {x - 2} = a \ge 0\\
\sqrt {y + 3} = b \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 4\\
{a^2} + {b^2} = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 4 - b\\
{\left( {4 - b} \right)^2} + {b^2} = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 4 - b\\
{b^2} - 4b + 4 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 4 - b\\
b = 2\left( {tm} \right)
\end{array} \right.\\
\Rightarrow a = b = 2\\
\Rightarrow \sqrt {x - 2} = \sqrt {y + 3} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x - 2 = 4\\
y + 3 = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 6\\
y = 1
\end{array} \right.
\end{array}$
