Đáp án:
\[x = y = z = \dfrac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\left( {x + y + z} \right) = \sqrt {4x - 1} + \sqrt {4y - 1} + \sqrt {4z - 1} \\
\Leftrightarrow 4\left( {x + y + z} \right) = 2\sqrt {4x - 1} + 2\sqrt {4y - 1} + 2\sqrt {4z - 1} \\
\Leftrightarrow \left[ {\left( {4x - 1} \right) - 2\sqrt {4x - 1} + 1} \right] + \left[ {\left( {4y - 1} \right) - 2\sqrt {4y - 1} + 1} \right] + \left[ {\left( {4z - 1} \right) - 2\sqrt {4z - 1} + 1} \right] = 0\\
\Leftrightarrow {\left( {\sqrt {4x - 1} - 1} \right)^2} + {\left( {\sqrt {4y - 1} - 1} \right)^2} + {\left( {\sqrt {4z - 1} - 1} \right)^2} = 0\,\,\,\,\,\,\,\left( 1 \right)\\
{\left( {\sqrt {4x - 1} - 1} \right)^2} \ge 0,\,\,\,\,\,\forall x\\
{\left( {\sqrt {4y - 1} - 1} \right)^2} \ge 0,\,\,\,\,\,\forall y\\
{\left( {\sqrt {4z - 1} - 1} \right)^2} \ge 0,\,\,\,\,\,\forall z\\
\Rightarrow {\left( {\sqrt {4x - 1} - 1} \right)^2} + {\left( {\sqrt {4y - 1} - 1} \right)^2} + {\left( {\sqrt {4z - 1} - 1} \right)^2} \ge 0,\,\,\,\forall x,y,z\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {\sqrt {4x - 1} - 1} \right)^2} = 0\\
{\left( {\sqrt {4y - 1} - 1} \right)^2} = 0\\
{\left( {\sqrt {4z - 1} - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {4x - 1} = 1\\
\sqrt {4y - 1} = 1\\
\sqrt {4z - 1} = 1
\end{array} \right. \Leftrightarrow x = y = z = \dfrac{1}{2}
\end{array}\)
Vậy \(x = y = z = \dfrac{1}{2}\)
