Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\left( {x > 0} \right)\\
9{x^{{{\log }_9}x}} = {x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
TH1:\,\,\,\,x = 1\\
\left( 1 \right) \Leftrightarrow 9 = 1\left( L \right)\\
TH2:\,\,\,0 < x \ne 1\\
\Leftrightarrow {\log _x}\left( {9{x^{{{\log }_9}x}}} \right) = {\log _x}{x^2}\\
\Leftrightarrow {\log _x}9 + {\log _x}{x^{{{\log }_9}x}} = 2\\
\Leftrightarrow {\log _x}9 + {\log _9}x = 2\\
\Leftrightarrow {\log _x}9 = 1\\
\Leftrightarrow x = 9\\
b,\,\,\,\,\,\left( {0 < x \ne 1} \right)\\
{x^4}{.5^3} = {5^{{{\log }_x}5}}\\
\Leftrightarrow {\log _5}\left( {{x^4}{{.5}^3}} \right) = {\log _5}\left( {{5^{{{\log }_x}5}}} \right)\\
\Leftrightarrow {\log _5}{x^4} + {\log _5}{5^3} = {\log _x}5\\
\Leftrightarrow 4{\log _5}x + 3 = {\log _x}5\\
\Leftrightarrow 4{\log _5}x - \frac{1}{{{{\log }_5}x}} + 3 = 0\\
\Leftrightarrow 4{\log _5}^2x + 3{\log _5}x - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\log _5}x = \frac{1}{4}\\
{\log _5}x = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt[4]{5}\\
x = \frac{1}{5}
\end{array} \right.
\end{array}\)
