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Điều kiện: $a > 0,\,a \ne 1,\,x > 0,\,x \ne 1$
Chọn $a$ làm cơ số :
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow \frac{{\log _a^2\left( {{\rm{ax}}} \right)}}{{{{\log }_a}x}} =
\frac{1}{2}{\log _a}\frac{1}{a} \Leftrightarrow {\left( {{{\log }_a}x + 1} \right)^2} = -
\frac{1}{2}{\log _a}x\\
\,\,\,\, \Leftrightarrow 2\log _a^2x + 5{\log _a}x + 2 = 0\Leftrightarrow \left[ \begin{array}{l}
{\log _a}x = - 2\\
{\log _a}x = - \frac{1}{2}
\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}
x = {a^{ - 2}}\\
x = {a^{ - \frac{1}{2}}}
\end{array} \right.
\end{array}$
Vậy $\left[ \begin{array}{l}
x = \frac{1}{{{a^2}}}\\
x = \frac{1}{{\sqrt a }}
\end{array} \right.$
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