Đáp án:
$\begin{array}{l}
1){\log _3}3x + 8 = 2 + x\left( {x > 0} \right)\\
\Leftrightarrow 1 + {\log _3}x + 8 = 2 + x\\
\Leftrightarrow x - {\log _3}x = 7\\
đặt\,{\log _3}x = t \Rightarrow x = {3^t}\\
pt \Rightarrow {3^t} - t = 7\\
Xét\,f\left( t \right) = {3^t} - t\\
f'\left( t \right) = {3^t}.\ln 3 - 1 > 0 \Rightarrow f\left( t \right)\,đồng\,biến\\
\Rightarrow {3^t} - t = 7\,có\,nghiệm\,duy\,nhất\,là\,t = 2\\
\Rightarrow x = {3^t} = {3^2} = 9\\
2)\log _2^3{\rm{[}}x\left( {x - 1} \right){\rm{]}} = 1\left( {dk:\left[ \begin{array}{l}
x > 1\\
x < 0
\end{array} \right.} \right)\\
\Leftrightarrow {\log _2}{\rm{[}}x\left( {x - 1} \right){\rm{]}} = 1\\
\Leftrightarrow x\left( {x - 1} \right) = 2\\
\Leftrightarrow {x^2} - x - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 1\left( {tm} \right)
\end{array} \right.
\end{array}$
