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\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=\left(1-1\right)+\left(1-1\right)-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(A=\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\left(\dfrac{1}{2006}-\dfrac{1}{2007}\right)-\left(\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
\(B=\dfrac{1}{2007}+\dfrac{1}{2009}-\dfrac{1}{2006}+\dfrac{1}{2008}\)
Dễ dàng thấy \(A>B\)
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(A=1+2+...+2^{2008}+2^{2009}\)
\(\Rightarrow2A=2+2^2+..+2^{2010}\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow S=1\)
S = 22010 - 22009 - 22008 - ... - 2 - 1
S= 22010 - ( 22009 + 22008 + ... + 2 + 1 )
Đặt A = 22009 + 22008 + .... + 2 + 1
2A = 2 . ( 22009 + 22008 + .... + 2 + 1
2A = 22010 + 22009 + .... + 22 + 2
2A - A = 22010 + 22009 + ...... + 22 + 2 - 22009 - 22008 - .... - 2 - 1
A = 22010 - 1
Thay A vào S ta có :
S = 22010 - ( 22010 - 1 )
S = 22010 - 22010 + 1
S = 0 + 1
S = 1
Vậy S = 1
a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b,\(7^{x+2}+2.7^{x-1}=345\)
\(7^{x-1}.\left(7^3+2\right)=345\)
\(\Rightarrow7^{x-1}.345=345\)
\(\Rightarrow7^{x-1}=345:345=1\)
\(\Rightarrow7^{x-1}=7^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)
\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)
\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
a,
\(\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}........\dfrac{-99}{100}.\dfrac{-120}{121}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.........\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}\)
\(=\dfrac{1.2.3.4.....10.3.4.5.6......11.12}{2^2.3^2........11^2}\)
\(=\dfrac{1.2.11.12}{2^2.11^2}=\dfrac{12}{22}\)
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\\ \Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(M=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2010}-M\)
* Tính M
\(M=2^{2009}+2^{2008}+...+2+1\\ \Rightarrow2^0+2^1+...+2^{2008}+2^{2009}\\ \Rightarrow2S=2^1+2^2+...+2^{2009}+2^{2010}\\ \Rightarrow2S-S=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\\ \Rightarrow S=2^{2010}-2^0=2^{2010}-1\)Thay M vào S, ta được :
\(S=2^{2010}-\left(2^{2010}-1\right)\\ \Rightarrow S=2^{2010}-2^{2010}+1\\ \Rightarrow S=1\)
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Rightarrow2S=2.\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)
\(\Rightarrow2S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)
Có \(2S-S=\left(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\right)-\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)
\(S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2-2^{2010}+2^{2009}+2^{2008}+...+2+1\)
\(S=2^{2011}+1\)
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