a . Ta có :
A = $2^{0}$+ $2^{1}$+$2^{2}$+ $2^{3}$+...+$2^{2010}$
⇒ 2A= $2^{1}$+$2^{2}$+ $2^{3}$+...+$2^{2010}$+$2^{2011}$
⇒ A = 2A-A=$2^{2011}$ - $2^{0}$ =$2^{2011}$ - 1 = B
→ A=B
b.Ta có A = 2009.2011=2009(2010+1)=2009.2010+2009
B = 20102 = 2010.2010=(2009+1)2010=2009.2010+2010
Vì 2010>2009 => 2009.2010+2009<2009.2010+2010 hay A<B
c. Ta có : A = $10^{30}$ =$(10³)^{10}$ = $1000^{10}$
B = $2^{100}$ =$(2^{10})^{10}$ =$1024^{10}$
Vì $1000^{10}$ < $1024^{10}$
=> A < B
@ara boss ris