giải chi tiết

Điều kiện: $\displaystyle 6^{x} -4^{x} \neq 0\Leftrightarrow x\neq 0$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{4.9^{x} -5.6^{x}}{6^{x} -4^{x}}  >3\Leftrightarrow \frac{4.\left(\frac{3}{2}\right)^{x} -5}{1-\left(\frac{2}{3}\right)^{x}}  >3\Leftrightarrow \begin{cases}
t=\left(\frac{2}{3}\right)^{x} & \\
\frac{4.\frac{1}{t} -5}{1-t}  >3 & ( 1)
\end{cases}\\
( 1) \Leftrightarrow \frac{4-5t}{t-t^{2}} -3 >0\\
\Leftrightarrow \frac{4-5t-3t+3t^{2}}{t( 1-t)}  >0\\
\Leftrightarrow \frac{3t^{2} -8t+4}{t( 1-t)}  >0\\
\Leftrightarrow \frac{( t-2)( 3t-2)}{t( 1-t)}  >0\\
\Leftrightarrow \left[ \begin{array}{l l}
0< t< \frac{2}{3} & \\
1< t< 2 & 
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l l}
\left(\frac{2}{3}\right)^{x} < \frac{2}{3} & \\
1< \left(\frac{2}{3}\right)^{x} < 2 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x >1 & \\
\log_{\frac{2}{3}} 2< x< 0 & 
\end{array} \right. \Rightarrow x\in (\log_{\frac{2}{3}} 2;0) \cup ( 1;+\infty )
\end{array}$
$\displaystyle \Rightarrow a=\frac{2}{3} ;\ b=2\Rightarrow a+b=\frac{8}{3} \Rightarrow A$