Đáp án:
\[x = 2\]
Giải thích các bước giải:
ĐKXĐ: \[x > \frac{{ - 5}}{2}\]
Ta có:
\[\begin{array}{l}
{\log _2}\left( {{x^2} - 3x + 6} \right) - {\log _3}\left( {2x + 5} \right) = 0\\
\Leftrightarrow {\log _2}\left( {{x^2} - 3x + 6} \right) = {\log _3}\left( {2x + 5} \right)
\end{array}\]
Đặt:\({\log _2}\left( {{x^2} - 3x + 6} \right) = {\log _3}\left( {2x + 5} \right) = t\)
Suy ra
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 3x + 6 = {2^t}\\
2x + 5 = {3^t}
\end{array} \right.\\
2x + 5 = {3^t}\\
\Leftrightarrow {\left( {2x + 5} \right)^2} = {\left( {{3^t}} \right)^2}\\
\Leftrightarrow 4{x^2} + 20x + 25 = {9^t}\\
\Leftrightarrow 4\left( {{x^2} - 3x + 6} \right) + 16\left( {2x + 5} \right) - 79 = {9^t}\\
\Leftrightarrow {4.2^t} + {16.3^t} = 79 + {9^t}\\
\Rightarrow t = 2\\
\Rightarrow x = 2
\end{array}\)
