Đáp án:
$A=1$
Giải thích các bước giải:
$A=2^{2012}-\bigg(\underbrace{2^{2011}+2^{2010}\ +\,.\!.\!.+\ 2+1}_{\large A'}\bigg)\\\Rightarrow A'=2^{2011}+2^{2010}\ +\,.\!.\!.+\ 2+1\\\Rightarrow2A'=2.(2^{2011}+2^{2010}\ +\,.\!.\!.+\ 2+1)\\\Rightarrow 2A'=2^{2012}+2^{2011}\ +\,.\!.\!.+\ 2^2+2\\\Rightarrow 2A'-A'=(2^{2012}+2^{2011}\ +\,.\!.\!.+\ 2^2+2)-(2^{2011}+2^{2010}\ +\,.\!.\!.+\ 2+1)\\\Rightarrow A'=2^{2012}-1\\\Rightarrow A=2^{2012}-(2^{2012}-1)\\\Rightarrow A=2^{2012}-2^{2012}+1\\\Rightarrow A=1$
Vậy $A=1$.
