Giải các bất phương trình :

$\begin{array}{l}
1)\,\,\,\,\,\,{\left( {{2^x} + {{3.2}^{ - x}}} \right)^{2{{\log }_2}x - {{\log }_2}\left( {x + 6} \right)}} > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,\,\,\,\,{\left( {{{4.3}^x} + {3^{ - x}}} \right)^{3{{\log }_3}\left( {x - 1} \right) - {{\log }_3}\left( {x - 1} \right)\left( {2x + 1} \right)}} > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array}$
1)
Điều kiện: x>0.
Khi đó lấy logarit cơ số 2 cho 2 vế dương ta có
$(1) \Leftrightarrow \left[ {2{{\log }_2}x - {{\log }_2}\left( {x + 6} \right)} \right].{\log _2}\left( {{2^x} + {{3.2}^{ - x}}} \right) > 0$
Từ $x > 0$$ \Rightarrow {2^x} > 1 \Rightarrow {2^x} + {3.2^{ - x}} > 1 \Rightarrow {\log _2}\left( {{2^x} + {{3.2}^{ - x}}} \right) > 0$
Do đó
$\begin{array}{l}
(1) \Leftrightarrow 2{\log _2}x - {\log _2}\left( {x + 6} \right) > 0 \\
     \Leftrightarrow {\log _2}{x^2} > {\log _2}\left( {x + 6} \right)\\
\,\,\,\,\,\,\, \Leftrightarrow {x^2} > x + 6\,\,\,\,\,\,\,\,\,\,\, \\
     \Leftrightarrow {x^2} - x - 6 > 0\\
\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x <  - 2\\
x > 3
\end{array} \right.
\end{array}$
Do $x > 0$ nên $x > 3$.
Vậy $x > 3$ là nghiệm thỏa mãn.

2)
Điều kiện: $x>1.$
Khi đó lấy logarit cơ  số 3 cho 2 vế ta có
$(2)\Leftrightarrow [3\log_3(x-1)-\log_3(x-1)-\log_3(2x+1)].\log_3(4.3^x+3^{-x})>0$
Do $x>1>0\Rightarrow \left\{ \begin{array}{l} 3^x>1\\ 3^{-x}>0 \end{array} \right.\Rightarrow \log_3(4.3^x+3^{-x})>0$
Suy ra
$(2)\Leftrightarrow \log_3(x-1)^3 -\log_3(x-1)-\log_3(2x+1)>0$
$\Leftrightarrow \log_3\frac{(x-1)^3}{(x-1)(2x+1)}>0$
$\Leftrightarrow  \frac{(x-1)^2}{(2x+1)} >1$
$\Leftrightarrow x^2-2x+1>2x+1$
$\Leftrightarrow x(x-4)>0\Leftrightarrow \left[\begin{array}{l} x>4\\ x<0 \end{array} \right.$
Kết hợp với điều kiện suy ra nghiệm BPT đã cho là: $x > 4$

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