Đáp án: $\,x = 3;y = 4;z = 6$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x \ge 2\\
y \ge 2\\
z \ge 5
\end{array} \right.\\
\sqrt {x - 2} + \sqrt {y - 3} + \sqrt {z - 5} = \frac{1}{2}\left( {x + y + z - 7} \right)\\
\Leftrightarrow x + y + z - 7 = 2\sqrt {x - 2} + 2\sqrt {y - 3} + 2\sqrt {z - 5} \\
\Leftrightarrow x - 2 - 2\sqrt {x - 2} + 1 + y - 3 - 2\sqrt {y - 3} + 1\\
+ z - 5 - 2\sqrt {z - 5} + 1 = 0\\
\Leftrightarrow {\left( {\sqrt {x - 2} - 1} \right)^2} + {\left( {\sqrt {y - 3} - 1} \right)^2} + {\left( {\sqrt {z - 5} - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x - 2} - 1 = 0\\
\sqrt {y - 3} - 1 = 0\\
\sqrt {z - 5} - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x - 2} = 1\\
\sqrt {y - 3} = 1\\
\sqrt {z - 5} = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2 = 1\\
y - 3 = 1\\
z - 5 = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3\left( {tm} \right)\\
y = 4\left( {tm} \right)\\
z = 6\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 3;y = 4;z = 6
\end{array}$
