M = 2008 + 2007/2 +2006/3 +...+2/2007 +1/2008
N = 1/2+1/3+1/4+1/5+...+1/2008+1/2009
tính M : N
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\(\frac{M}{N}=\frac{\frac{1}{2007}+\frac{2}{2006}+......+\frac{2006}{2}+\frac{2007}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2006}+\frac{1}{2007}}\)
\(\frac{M}{N}=\frac{\frac{1}{2007}+1+\frac{2}{2006}+1+.......+\frac{2007}{1}+1+\frac{2008}{2008}-2008}{\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+.....+\frac{1}{2}}\)
\(\frac{M}{N}=\frac{\frac{2008}{2007}+\frac{2008}{2006}+....+\frac{2008}{1}+\frac{2008}{2008}-2008}{\frac{1}{2008}+........+\frac{1}{2}}\)
đến đây là ra rùi ha
\(B=2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=1+1+\frac{2007}{2}+1+\frac{2006}{3}+...+1+\frac{1}{2008}\)
\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)
\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)
Suy ra \(A=2009\).
A=2008+2007/2+2006/3+2005/4+...+2/2007+1/2008
1/2+1/3+1/4+1/5+...+1/2007+1/2008
=(1+2007/2)+(1+2006/3)+(1+2005/4)+...+(1+2/2007)+(1+1/2008)
1/2+1/3+1/4+...+1/2008
=2009(1/2+1/3+1/4+...+1/2008)
1/2+1/3+1/4+..+1/2008
=2009
\(M:N=\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Ta có tử số bằng: 2008+2007/2+2006/3+2005/4+…..+2/2007+1/2008
(Phân tích 2008 thành 2008 con số 1 rồi đưa vào các nhóm)
= (1 + 2007/2) + (1 + 2006/3) + (1 + 2005/4) +... + (1 + 2/2007) + ( 1 + 1/2008) + (1)
= 2009/2 + 2009/3 + 2009//4 + ……. + 2009/2007 + 2009/2008 + 2009/2009
= 2009 x (1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009)
Mẫu số: 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009
\(\Rightarrow M:N=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)