Đáp án: $\,\left( {x;y} \right) = \left\{ {\left( {2;22} \right);\left( {26; - 2} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 1;y \ge - 3\\
\left\{ \begin{array}{l}
\sqrt {x - 1} + \sqrt {y + 3} = 6\\
x + y = 24
\end{array} \right.\\
\Leftrightarrow x - 1 + 2\sqrt {x - 1} .\sqrt {y + 3} + y + 3 = 36\\
\Leftrightarrow x + y + 2 + 2\sqrt {\left( {x - 1} \right)\left( {y + 3} \right)} = 36\\
\Leftrightarrow 24 + 2 + 2\sqrt {\left( {x - 1} \right)\left( {y + 3} \right)} = 36\\
\Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {y + 3} \right)} = 5\\
\Leftrightarrow \left( {x - 1} \right)\left( {y + 3} \right) = 25\\
\Leftrightarrow \left( {24 - y - 1} \right)\left( {y + 3} \right) = 25\left( {do:x + y = 24} \right)\\
\Leftrightarrow \left( {23 - y} \right)\left( {y + 3} \right) = 25\\
\Leftrightarrow - {y^2} + 20y + 69 = 25\\
\Leftrightarrow {y^2} - 20y - 44 = 0\\
\Leftrightarrow \left( {y - 22} \right)\left( {y + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = 22\left( {tmdk} \right) \Leftrightarrow x = 2\left( {tm} \right)\\
y = - 2\left( {tmdk} \right) \Leftrightarrow x = 26\left( {tm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;22} \right);\left( {26; - 2} \right)} \right\}
\end{array}$
