Giải bất phương trình :
  $\log _3\sqrt {x^2 - 5x + 6}  + \log _\frac{1}{3}\sqrt {x - 2}  > \frac{1}{2}{\log _\frac{1}{3}}\left( {x + 3} \right)        (1)$
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}$
$\begin{array}{l}
(1) \Leftrightarrow {\log _3}\sqrt {\left( {x - 2} \right)\left( {x - 3} \right)}  - {\log _3}\sqrt {x - 2}  >  - {\log _3}\sqrt {x + 3} \\
\,\,\,\,\,\,\, \Leftrightarrow {\log _3}\sqrt {x - 3}  >  - {\log _3}\sqrt {x + 3} \\
\,\,\,\,\,\,\, \Leftrightarrow {\log _3}\sqrt {{x^2} - 9}  > 0\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \sqrt {{x^2} - 9}  > 1\\
\,\,\,\,\,\,\, \Leftrightarrow {x^{\left[ 2 \right.}} > 10\\
\,\,\,\,\,\,\, \Leftrightarrow x > \sqrt {10}
\end{array}$
cám ơn bạn nhé ^_^ –  anh_chang_zaizai_90 11-07-12 01:01 PM

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