Đáp án:
$\begin{array}{l}
Dkxd:\left\{ {x;y;z \ge \dfrac{1}{4}} \right.\\
\left\{ \begin{array}{l}
x + y = \sqrt {4z - 1} \left( 1 \right)\\
y + z = \sqrt {4x - 1} \left( 2 \right)\\
z + x = \sqrt {4y - 1}
\end{array} \right.\\
\left( 1 \right) - \left( 2 \right)\\
\Rightarrow x + y - y - z = \sqrt {4z - 1} - \sqrt {4x - 1} \\
\Rightarrow x - z = \dfrac{{4z - 1 - \left( {4x - 1} \right)}}{{\sqrt {4z - 1} + \sqrt {4x - 1} }}\\
\Rightarrow x - z = \dfrac{{ - 4\left( {x - z} \right)}}{{\sqrt {4z - 1} + \sqrt {4x - 1} }}\\
\Rightarrow \left( {x - z} \right).\left( {1 + \dfrac{4}{{\sqrt {4z - 1} + \sqrt {4x - 1} }}} \right) = 0\\
\Rightarrow x - z = 0\\
\Rightarrow x = z\\
TT:x = y = z\\
\left( 1 \right) \Rightarrow x + x = \sqrt {4x - 1} \\
\Rightarrow 2x = \sqrt {4x - 1} \\
\Rightarrow 4{x^2} = 4x - 1\\
\Rightarrow 4{x^2} - 4x + 1 = 0\\
\Rightarrow {\left( {2x - 1} \right)^2} = 0\\
\Rightarrow x = \dfrac{1}{2}\left( {tmdk} \right)\\
Vậy\,x = y = z = \dfrac{1}{2}
\end{array}$
