Đáp án:
$\begin{array}{l}
Dkxd:x \ge 1;y \ge 2;z \ge 3\\
x + y + z + 8 = 2\sqrt {x - 1} + 4\sqrt {y - 2} + 6\sqrt {z - 3} \\
\Rightarrow x - 1 - 2\sqrt {x - 1} + 1\\
+ y - 2 - 4\sqrt {y - 2} + 4\\
+ z - 3 - 6\sqrt {z - 3} + 9 = 0\\
\Rightarrow {\left( {\sqrt {x - 1} - 1} \right)^2} + {\left( {\sqrt {y - 2} - 2} \right)^2} + {\left( {\sqrt {z - 3} - 3} \right)^2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 1} = 1\\
\sqrt {y - 2} = 2\\
\sqrt {z - 3} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = 1\\
y - 2 = 4\\
z - 3 = 9
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 6\\
z = 12
\end{array} \right.\left( {tmdk} \right)\\
Vay\,\left( {x;y;z} \right) = \left( {2;6;12} \right)
\end{array}$
