Đáp án: $x = \frac{{ - 7 + \sqrt {41} }}{2}$
Giải thích các bước giải:
$\begin{array}{l}
{\log _2}\left( {x + 2} \right) + {\log _4}{\left( {x + 5} \right)^2} + {\log _{\frac{1}{2}}}8 = 0\left( {dk:x \ge - 2} \right)\\
\Rightarrow {\log _2}\left( {x + 2} \right) + {\log _2}\left| {x + 5} \right| - {\log _2}8 = 0\\
\Rightarrow {\log _2}\frac{{\left( {x + 2} \right).\left| {x + 5} \right|}}{8} = 0\\
\Rightarrow \frac{{\left( {x + 2} \right).\left| {x + 5} \right|}}{8} = 1\\
\Rightarrow \left( {x + 2} \right).\left| {x + 5} \right| = 8\left( * \right)\\
+ )Khi:x \ge - 5\\
\left( * \right) \Rightarrow \left( {x + 2} \right)\left( {x + 5} \right) = 8\\
\Rightarrow {x^2} + 7x + 10 - 8 = 0\\
\Rightarrow {x^2} + 7x + 2 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{ - 7 + \sqrt {41} }}{2}\left( {tm} \right)\\
x = \frac{{ - 7 - \sqrt {41} }}{2}\left( {ktm} \right)
\end{array} \right.\\
+ )Khi:x < - 5\\
\left( * \right) \Rightarrow - \left( {x + 2} \right)\left( {x + 5} \right) = 8\\
\Rightarrow {x^2} + 7x + 10 + 8 = 0\\
\Rightarrow {x^2} + 7x + 18 = 0\\
\Rightarrow x \in \emptyset
\end{array}$
