Đáp án:
$\begin{array}{l}
{\left( {{{\log }_3}x} \right)^2} - 2{\log _3}x = - 1\\
\Rightarrow {\left( {{{\log }_3}x} \right)^2} - 2{\log _3}x + 1 = 0\\
\Rightarrow {\log _3}x = 1\\
B = 3{\log _{\sqrt 3 }}\sqrt x - 6{\log _9}\left( {3x} \right) + {\log _{\dfrac{1}{3}}}\left( {\dfrac{x}{9}} \right)\\
= 3{\log _3}x - 6.{\log _{{3^2}}}\left( {3x} \right) - {\log _3}\left( {\dfrac{x}{{{3^2}}}} \right)\\
= 3{\log _3}x - 6.\dfrac{1}{2}{\log _3}\left( {3x} \right) - \left( {{{\log }_3}x - {{\log }_3}{3^2}} \right)\\
= 3{\log _3}x - 3\left( {{{\log }_3}3 + {{\log }_3}x} \right) - {\log _3}x + 2\\
= 3{\log _3}x - 3 - 3{\log _3}x - {\log _3}x + 2\\
= - 1 - {\log _3}x\\
= - 1 - 1\\
= - 2
\end{array}$
