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Điều kiện : $\left\{ \begin{array}{l}
{x^2} - 5x + 6 > 0\\
x - 2 > 0\\
x + 3 > 0
\end{array} \right. \Leftrightarrow x > 3$
$\begin{array}{l}
{\log
_3}\sqrt {{x^2} - 5x + 6} + {\log _{\frac{1}{3}}}\sqrt {x - 2} >
\frac{1}{2}{\log _{\frac{1}{3}}}\left( {x + 3} \right)\, (1)\\
{x^2}{\log _x}27.{\log _9}x > x + 4 (1)\\
\end{array}$
$\begin{array}{l}
(1)
\Leftrightarrow {\log _3}\sqrt {\left( {x - 2} \right)\left( {x - 3}
\right)} - {\log _3}\sqrt {x - 2} > - {\log _3}\sqrt {x + 3} \\
\,\,\,\,\,\,\, \Leftrightarrow {\log _3}\sqrt {x - 3} > - {\log _3}\sqrt {x + 3} \\
\,\,\,\,\,\,\,
\Leftrightarrow {\log _3}\sqrt {{x^2} - 9} >
0\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \sqrt {{x^2} - 9} > 1\\
\,\,\,\,\,\,\, \Leftrightarrow {x^{\left[ 2 \right.}} > 10\\
\,\,\,\,\,\,\, \Leftrightarrow x > \sqrt {10}
\end{array}$
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Trả lời 11-07-12 12:48 PM
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