a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b,\(7^{x+2}+2.7^{x-1}=345\)
\(7^{x-1}.\left(7^3+2\right)=345\)
\(\Rightarrow7^{x-1}.345=345\)
\(\Rightarrow7^{x-1}=345:345=1\)
\(\Rightarrow7^{x-1}=7^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy \(x=1\)