2^2010-(2^2009+2^2008+2^2007+......+2^1+2^0)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
M=2^2010-(2^2009+2^2008+2^2007+...+2^1+2^0)
M=22010-22009-22008-22007-...-21-20
=>2M=22011-22010-22009-22008-...-22-21
=>2M-M=22011-22010-22009-22008-...-22-21-(22010-22009-22008-22007-...-21-20)
=>M=22011-22010-22009-22008-...-22-21-22010+22009+22008+22007+...+21+20
=22011-22010-22010+20
=22011-2.22010+1
=22011-22011+1
=1
vậy M=1
đúng mjk với nha
ĐẶt A = 2^0 + 2^1 +.. + 2^2009
2A = 2^ 1 + 2^2 +.... + 2^2009 +2 ^2010
2A - A = 2^1 + 2^2 + . ... + 2^2009 +2^2010 - 2 ^0 - 2^1 - 2^2 -..-2^3009
A = 2^2010 - 2^0 = 2^2010 - 1
M = 2^2010 - A = 2^2010 - (2^2010 - 1) = 2^2010 - 2^2010 +1 = 1
M = 22010 - ( 22009 + 22008 + 22007 + ... + 21 + 20)
Ta có: N = 22009 + 22008 + 22007 + ... + 21 + 20
2N = 22010 + 22009 + 22008 + ... + 22 + 21
N = 2N - N = ( 22010 + 22009 + 22008 + ... + 22 + 21) - ( 22009 + 22008 + 22007 + ... + 21 + 20 )
N = 22010 - 20
Suy ra: M = 22010 - 22010 - 20
M = 0 - 1
M = -1
Vậy M = -1
M=2^2010-(2^2009+2^2008+2^2007+...+2^1+2^0)
M=22010-22009-22008-22007-...-21-20
=>2M=22011-22010-22009-22008-...-22-21
=>2M-M=22011-22010-22009-22008-...-22-21-(22010-22009-22008-22007-...-21-20)
=>M=22011-22010-22009-22008-...-22-21-22010+22009+22008+22007+...+21+20
=22011-22010-22010+20
=22011-2.22010+1
=22011-22011+1
=1
Vậy........
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..
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