2^2010-(2^2009+2^2008+...+2^1+2^0)
TA
Những câu hỏi liên quan
a) 2010/1+2009/2+2008/3+ ... +1/2010+2010 : 1+1/2+1/3+ ... +1/2010=
b) 1/2011+1/2010+1/2009+ ... +1/3+1/2 : 2010/1+2009/2+2008/3+ ... +1/2010=
Tính: 2^2010 - (2^2009 + 2^2008+ .....+ 2^1 + 2^0)
Đặt A = 22009 + 22008 + ... + 21 + 20
2A = 22010 + 22009 + ... + 22 + 21
2A - A = (22010 + 22009 + ... + 22 + 21) - (22009 + 22008 + ... + 21 + 20)
A = 22010 - 20
A = 22010 - 1
=> 22010 - (22009 + 22008 + ... + 21 + 20)
= 22010 - (22010 - 1)
= 22010 - 22010 + 1
= 1
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Đặt A = 22009 + 22008 + ... + 21 + 20
2A = 22010 + 22009 + ... + 22 + 21
2A - A = (22010 + 22009 + ... + 22 + 21) - (22009 + 22008 + ... + 21 + 20)
A = 22010 - 20
A = 22010 - 1
=> 22010 - (22009 + 22008 + ... + 21 + 20)
= 22010 - (22010 - 1)
= 22010 - 22010 + 1
= 1
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M=2^2010-(2^2009+2^2008+...+2^1+2^0)
tính 2010*2010-2009*2009+2008*2008-........+2*2-1*1
Tìm m = 2^2010 - ( 2^2009 + 2^2008 + .... + 2^1 + 2^0 )
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=1+2+2^2+...+2^{2008}+2^{2009}\)
\(2\left(2^{2010}-M\right)=2+2^2+2^3+...+2^{2009}+2^{2010}\)
\(2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+2^3+...+2^{2009}+2^{2010}\right)-\left(1+2+2^2+...+2^{2008}+2^{2009}\right)\)
\(2^{2010}-M=2^{2010}-1\)
\(M=2^{2010}-2^{2010}+1\)
\(M=1\)
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Đặt \(M=2^{2010}-A\)
Ta có:
\(A=2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow M=\left(2^{2010}-2^{2010}\right)+1\)
\(\Rightarrow M=1\)
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Tính: M= 2^2010-( 2^2009 + 2^2008+....+2^1 +2^0)
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
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T=2^2010-(2^2009+2^2008+...+2^1+2^0)
\(T=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(T=2^{2010}-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\)
Đặt: \(A=2^0+2^1+....+2^{2008}+2^{2009}\)
\(2A=2\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\)
\(2A=2^1+2^2+....+2^{2009}+2^{2010}\)
\(2A-A=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+....+2^{2008}+2^{2009}\right)\)\(A=2^{2010}-1\)
Thay \(A\) vào \(T\) ta có:
\(T=2^{2010}-2^{2010}+1=1\)
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Tìm x thỏa mãn: x + (x + 1) + (x + 2) + … + 2009 + 2010 = 2010 A.-2010 B.-2008 C.0 D.-2009
2^2010-(2^2009+2^2008+2^2007+......+2^1+2^0)
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..
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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
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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
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