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NT
1
HT
12 tháng 9 2015
Đặt N = 22009 + 22008 + 22007 +......+ 21 + 20
2N = 22010 + 22009 + 22008 +.....+ 22 + 21
2N - N = 22010 - 20
=> N = 22010 - 1
=> M = 22010 - (22010 - 1)
=> M = 22010 - 22010 + 1
=> M = 1
AM
5 tháng 7 2015
Đặt N=22009+22008+...+1
=>2N=22010+22009+...+2
=>2N-N=(22010+22009+...+2)-(22009+22008+...+1)
=>N=22010-1
Mà M=22010-N=22010-(22010-1)=1
27 tháng 5 2022
Đặt \(N=2^0+2^1+...+2^{2008}+2^{2009}\)
Suy ra: \(M=2^{2010}-N\)
Ta có: \(N=2^0+2^1+...+2^{2008}+2^{2009}\)
\(\Leftrightarrow2N=2+2^2+...+2^{2009}+2^{2010}\)
\(\Leftrightarrow N=2^{2010}-1\)
\(M=2^{2010}-N=2^{2010}-2^{2010}+1=1\)
PN
1
31 tháng 5 2017
Đặt \(A=2^{2009}+2^{2008}+...+2+2^0\)
\(=1+2+...+2^{2008}+2^{2009}\)
\(\Rightarrow2A=2+2^2+...+2^{2010}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{2010}\right)-\left(1+2+...+2^{2009}\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(=2^{2010}-2^{2010}+1=1\)
Vậy M = 1
MV
5 tháng 7 2017
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
Gọi \(N=2^{2009}+2^{2008}+...+2^1+2^0\)
\(2N=2^{2010}+2^{2009}+...+2^2+2^1\\ 2N-N=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\\ N=2^{2010}-2^0\\ N=2^{2010}-1\)
Thay vào ta được
\(M=2^{2010}-\left(2^{2010}-1\right)\\ M=2^{2010}-2^{2010}+1\\ M=1\)
Vậy \(M=1\)
5 tháng 7 2017
Ta có :
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^0\right)\)
Đặt A=22009+22008+..+20
\(A=2^{2009}+2^{2008}+...+2^0\\ 2A=2^{2010}+2^{2009}+...+2^1\\ \Rightarrow2A-A=A=2^{2010}-2^0\\ \Rightarrow M=2^{2010}-\left(2^{2010}-2^0\right)\\ M=2^{2010}-2^{2010}+1\\ \Rightarrow M=1\)
Chúc bạn học tốt!
NT
2
8 tháng 8 2019
Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)
Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)
Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)
KM
8 tháng 8 2019
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)
\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)
\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)
\(2^{2010}-M=2^{2010}-1\)
=> M = 1
MA
14 tháng 9 2016
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=2^0+2^1+...+2^{2008}+2^{2009}\)
\(2\left(2^{2010}-M\right)=2+2^2+...+2^{2009}+2^{2010}\)
\(2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\)
\(2^{2010}-M=2^{2010}-1\)
\(M=2^{2010}-2^{2010}+1\)
\(M=1\)
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=1+2+2^2+...+2^{2008}+2^{2009}\)
\(2\left(2^{2010}-M\right)=2+2^2+2^3+...+2^{2009}+2^{2010}\)
\(2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+2^3+...+2^{2009}+2^{2010}\right)-\left(1+2+2^2+...+2^{2008}+2^{2009}\right)\)
\(2^{2010}-M=2^{2010}-1\)
\(M=2^{2010}-2^{2010}+1\)
\(M=1\)
Đặt \(M=2^{2010}-A\)
Ta có:
\(A=2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow M=\left(2^{2010}-2^{2010}\right)+1\)
\(\Rightarrow M=1\)