tính C=2010/1+2009/2 + 2008/3 +...+ 1/2010:1\2+1/3+1/4+......+1\2011
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Ta có: \(C=\frac{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)
Đặt \(A=\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}\)
\(A=\frac{2010}{1}+1+\frac{2009}{1}+1+\frac{2008}{1}+1+...+\frac{1}{2010}+1-2010\)
\(=\frac{2011}{1}+\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}-\frac{2011.2010}{2011}\)
\(=2011\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-1\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\)
Ta có: \(C=\frac{A}{B}=2011\)(lấy A-B)
Ta có :
\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=\dfrac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)
\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=\dfrac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)
Vì \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)
~ Học tốt ~
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)
\(A=\frac{1}{2011}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+....+\frac{1}{2010}}\)
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x 0
=0
*2010/1+2009/2+...+1/2010
=(2009/2+1)+(2008/3+1)+...+(1/2010+1)+1
=2011/2+2011/3+..+2011/2010+2011/2011
=2011(1/2+1/3+1/4+...+1/2011)
=> C=2011/1=2011