2008-1/2008=2007/2008

1/2-1/2009=2007/2009

\(B=2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}\)

\(=1+1+\frac{2007}{2}+1+\frac{2006}{3}+...+1+\frac{1}{2008}\)

\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)

\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)

Suy ra \(A=2009\).

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot......\cdot\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot......\cdot\frac{19}{20}\)

\(A=\frac{1.2.3.....19}{2.3........20}\)

\(A=\frac{1}{20}\)

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

Kết quả bằng 2009

Xét tử

2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1

=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008

=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+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
Vậy  A = 2009