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\(A=2^{2010}+2^{2009}+...+2^2+2\)
\(\Rightarrow2A=2^{2011}+2^{2010}+...+2^3+2^2\)
\(\Rightarrow2A-A=\left(2^{2011}+2^{2010}+...+2^3+2^2\right)-\left(2^{2010}+2^{2009}+...+2^2+2\right)\)
\(\Rightarrow A=2^{2011}-2\)
Vậy \(A=2^{2011}-2\)
\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
\(\Leftrightarrow x=2012\)
1.
M = 22010 - ( 22009 + 22008 + ... + 21 + 20 )
đặt N = 22009 + 22008 + ... + 21 + 20
2N = 22010 + 22009 + ... + 22 + 21
2N - N = ( 22010 + 22009 + ... + 22 + 21 ) - ( 22009 + 22008 + ... + 21 + 20 )
N = 22010 - 20
Thay N vào ta được :
M = 22010 - ( 22010 - 20 )
M = 22010 - 22010 + 20
M = 20 = 1
2.
Ta có :
2332 < 2333 = ( 23 ) 111 = 8111
3223 > 3222 = ( 32 ) 111 = 9111
Vì 2332 < 8111 < 9111 < 3223
Ta có ;
\(P=2^{2010}-2^{2009}-2^{2008}-..............-2-1\)
\(\Leftrightarrow P=2^{2010}-\left(2^{2009}+2^{2008}+..........+2+1\right)\)
Đặt :
\(A=2^{2009}+2^{2008}+..........+2+1\)
\(\Leftrightarrow2A=2^{2010}+2^{2009}+...........+2\)
\(\Leftrightarrow2A-A=\left(2^{2010}+2^{2009}+.......+2\right)-\left(2^{2009}+2^{2008}+........+2+1\right)\)
\(\Leftrightarrow A=2^{2010}-1\)
\(\Leftrightarrow P=2^{2010}-\left(2^{2010}-1\right)\)
\(\Leftrightarrow P=2^{2010}-2^{2010}+1\)
\(\Leftrightarrow P=0+1=1\)