Question

2^2010-(2^2009+2^2008+2^2007+......+2^1+2^0)

Answer 1

Đặt :

\(A=2^{2009}+2^{2008}+......+2+1\)

\(\Leftrightarrow2A=2^{2010}+2^{2009}+......+2^2+2\)

\(\Leftrightarrow2A-A=\left(2^{2010}+2^{2009}+.....+2\right)-\left(2^{2009}+2^{2008}+.....+2+1\right)\)

\(\Leftrightarrow A=2^{2010}-1\)

\(\Leftrightarrow2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=2^{2010}-2^{2010}+1=1\)

Vậy..

Answer 2

Đặt A=\(2^{2010}-\left(2^{2009}+2^{2008}+2^{2007}+...+2^1+2^0\right)\)

Khi đó:\(A=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\\ \Rightarrow2A=2^{2011}-2^{2010}-2^{2009}-...-2^1\\ 2A-A=2^{2011}-2^{2010}-2^{2009}-...-2^1-\left(2^{2010}-2^{2009}-....-2^1-2^0\right)\\ A=2^{2011}-2^{2010}-...-2^1+2^{2010}+2^{2009}+...+2^0\\ A=2^{2011}-2.2^{2010}+2^0\\ A=1\)Vậy A=1

Answer 3

Gọi M=\(2^{2010}-\left(2^{2009}+2^{2008}....+2^1+2^0\right)\)

\(2^{2010}-\)M=\(2^{2010}-2^{2009}-2^{2008}-2^{2007}-....2^1-2^0\)

2.(\(2^{2010}\)-M)=\(2^1+2^2+....+2^{2009}+2^{2010}\)

2.(\(2^{2010}\)-M)-(\(2^{2010}\) -M)=(\(2^1+2^2+....+2^{2009}+2^{2010}\))-(\(2^0+2^1+....+2^{2008}+2^{2009}\)

\(2^{2010}-\)M=\(2^{2010}-1\) M=\(2^{2010}-2^{2010}+1\) M=1