Đáp án:
$\left( {x;y;z} \right) = \left( {0;4;2} \right)$
Giải thích các bước giải:
$\begin{array}{l}
c)\sqrt {x + 1} + \sqrt {y - 3} + \sqrt {z - 1} = \dfrac{1}{2}\left( {x + y + z} \right)\left( {DK:x \ge - 1;y \ge 3;z \ge 1} \right)\\
\Leftrightarrow x + y + z - 2\sqrt {x + 1} - 2\sqrt {y - 3} - 2\sqrt {z - 1} = 0\\
\Leftrightarrow \left( {x + 1 - 2\sqrt {x + 1} + 1} \right) + \left( {y - 3 - 2\sqrt {y - 3} + 1} \right) + \left( {z - 1 - 2\sqrt {z - 1} + 1} \right) = 0\\
\Leftrightarrow {\left( {\sqrt {x + 1} - 1} \right)^2} + {\left( {\sqrt {y - 3} - 1} \right)^2} + {\left( {\sqrt {z - 1} - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x + 1} - 1 = 0\\
\sqrt {y - 3} - 1 = 0\\
\sqrt {z - 1} - 1 = 0
\end{array} \right.\left( {Do:{{\left( {\sqrt {x + 1} - 1} \right)}^2} + {{\left( {\sqrt {y - 3} - 1} \right)}^2} + {{\left( {\sqrt {z - 1} - 1} \right)}^2} \ge 0,\forall x \ge - 1;y \ge 3;z \ge 1} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 1\\
y - 3 = 1\\
z - 1 = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 4\\
z = 2
\end{array} \right.
\end{array}$
Vậy $\left( {x;y;z} \right) = \left( {0;4;2} \right)$
