Đáp án:
$\begin{array}{l}
{3^{\log _2^2x}} + {x^{\log _2^23}} = 6\left( {dk:x > 0} \right)\\
\Leftrightarrow {\left( {{3^{{{\log }_2}x}}} \right)^2} + {\left( {{3^{{{\log }_2}x}}} \right)^2} = 6\\
\Leftrightarrow 2.{\left( {{3^{{{\log }_2}x}}} \right)^2} = 6\\
\Leftrightarrow {\left( {{3^{{{\log }_2}x}}} \right)^2} = 3\\
\Leftrightarrow {3^{{{\log }_2}x}} = {3^{\frac{1}{2}}}\left( {do:{3^{{{\log }_2}x}} > 0} \right)\\
\Leftrightarrow {\log _2}x = \frac{1}{2}\\
\Leftrightarrow x = {2^{\frac{1}{2}}} = \sqrt 2 \left( {tm} \right)\\
Vậy\,x = \sqrt 2
\end{array}$
