Đáp án:
$\begin{array}{l}
+ ){\log _2}\left( {x + 2} \right) = {\log _4}\left( {8x + 4} \right)\left( {x > - 2} \right)\\
\Rightarrow {\log _2}\left( {x + 2} \right) = {\log _{{2^2}}}\left[ {4\left( {x + 2} \right)} \right]\\
\Rightarrow {\log _2}\left( {x + 2} \right) = 1 + \frac{1}{2}{\log _2}\left( {x + 2} \right)\\
\Rightarrow \frac{1}{2}{\log _2}\left( {x + 2} \right) = 1\\
\Rightarrow {\log _2}\left( {x + 2} \right) = 2\\
\Rightarrow x + 2 = 4\\
\Rightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2\\
+ ){\log _{\frac{1}{3}}}\left( {{4^x} + 4} \right) \le {\log _{\frac{1}{3}}}\left( {{2^{2x + 1}} - {{3.2}^x}} \right)\\
\left( {dk:{{2.2}^x} > 3 \Rightarrow x > {{\log }_2}\frac{3}{2}} \right)\\
Pt \Rightarrow {2^{2x}} + 4 \ge {2.2^{2x}} - {3.2^x}\\
\Rightarrow {2^{2x}} - {3.2^x} - 4 \le 0\\
\Rightarrow - 1 \le {2^x} \le 4\\
\Rightarrow 0 < {2^x} \le {2^2}\\
\Rightarrow x \le 2\\
Vậy\,{\log _2}\frac{3}{2} < x \le 2\\
+ )\log _5^2x - 3{\log _5}x + 2 \le 0\left( {x > 0} \right)\\
\Rightarrow 1 \le {\log _5}x \le 2\\
\Rightarrow 5 \le x \le 25
\end{array}$
