Đáp án:
\[x = y = z = 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y - 2 = 2\sqrt {2z - 3} \\
y + z - 2 = 2\sqrt {2x - 3} \\
z + x - 2 = 2\sqrt {2y - 3}
\end{array} \right.\\
\Rightarrow \left( {x + y - 2} \right) + \left( {y + z - 2} \right) + \left( {z + x - 2} \right) = 2\sqrt {2z - 3} + 2\sqrt {2x - 3} + 2\sqrt {2y - 3} \\
\Leftrightarrow 2x + 2y + 2z - 6 - 2\sqrt {2z - 3} - 2\sqrt {2x - 3} - 2\sqrt {2y - 3} = 0\\
\Leftrightarrow \left[ {\left( {2x - 3} \right) - 2\sqrt {2x - 3} + 1} \right] + \left[ {\left( {2y - 3} \right) - 2\sqrt {2y - 3} + 1} \right] + \left[ {\left( {2z - 3} \right) - 2\sqrt {2z - 3} + 1} \right] = 0\\
\Leftrightarrow {\left( {\sqrt {2x - 3} - 1} \right)^2} + {\left( {\sqrt {2y - 3} - 1} \right)^2} + {\left( {\sqrt {2z - 3} - 1} \right)^2} = 0\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
{\left( {\sqrt {2x - 3} - 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {\sqrt {2y - 3} - 1} \right)^2} \ge 0,\,\,\,\forall y\\
{\left( {\sqrt {2z - 3} - 1} \right)^2} \ge 0,\,\,\,\forall z\\
\Rightarrow {\left( {\sqrt {2x - 3} - 1} \right)^2} + {\left( {\sqrt {2y - 3} - 1} \right)^2} + {\left( {\sqrt {2z - 3} - 1} \right)^2} \ge 0,\,\,\,\forall x,y,z\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {\sqrt {2x - 3} - 1} \right)^2} = 0\\
{\left( {\sqrt {2y - 3} - 1} \right)^2} = 0\\
{\left( {\sqrt {2z - 3} - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {2x - 3} = 1\\
\sqrt {2y - 3} = 1\\
\sqrt {2z - 3} = 1
\end{array} \right. \Leftrightarrow x = y = z = 2
\end{array}\)
Vậy \(x = y = z = 2\)
