Số nghiệm của phương trình \({\log _2}\left( {x + 2} \right) + {\log _4}{\left( {x - 5} \right)^2} + {\log _{\frac{1}{2}}}8 = 0\) là

Đáp án A

Điều kiện: \( - 2 < x \ne 5\).

\({\log _2}\left( {x + 2} \right) + {\log _4}{\left( {x - 5} \right)^2} + {\log _{\frac{1}{2}}}8 = 0 \Leftrightarrow {\log _2}\left( {x + 2} \right) + {\log _2}\left| {x - 5} \right| = {\log _2}8\)

\( \Leftrightarrow {\log _2}\left[ {\left( {x + 2} \right)\left| {x - 5} \right|} \right] = {\log _2}8 \Leftrightarrow \left( {x + 2} \right)\left| {x - 5} \right| = 8\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 5\\\left( {x + 2} \right)\left( {x - 5} \right) = 8\end{array} \right.\\\left\{ \begin{array}{l} - 2 < x < 5\\\left( {x + 2} \right)\left( {5 - x} \right) = 8\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 5\\{x^2} - 3{\rm{x}} - 18 = 0\end{array} \right.\\\left\{ \begin{array}{l} - 2 < x < 5\\ - {x^2} + 3{\rm{x}} + 2 = 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 5\\x = 6 \vee x = - 3\end{array} \right.\\\left\{ \begin{array}{l} - 2 < x < 5\\x = \frac{{3 + \sqrt {17} }}{2} \vee x = \frac{{3 - \sqrt {17} }}{2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 6\\x = \frac{{3 + \sqrt {17} }}{2}\\x = \frac{{3 - \sqrt {17} }}{2}\end{array} \right.\).