(x - 5).(x - 7) = 0

=> \(\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x=7\end{cases}}\)

Vậy \(x\in\left\{5;7\right\}\)

\(A=1+2+2^2+2^3+.....+2^{2008}\)

\(\Rightarrow2A=2+2^2+2^3+......+2^{2009}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+......+2^{2009}\right)-\left(1+2+2^2+.....+2^{2008}\right)\)

\(\Rightarrow A=2^{2009}-1\)

\(\Rightarrow A-B=2^{2009}-1-2^{2009}=-1\)