5/
A = $\frac{-8}{1.5}$ + $\frac{-8}{5.9}$ + .... + $\frac{-8}{2001.2005}$
A = -8 . ($\frac{1}{1,5}$ + $\frac{1}{5,9}$ + ... + $\frac{1}{2001.2005}$)
A = $\frac{-8}{4}$ . ($\frac{4}{1.5}$ + $\frac{4}{5.9}$ + ... + $\frac{4}{2001.2005}$)
A = -2 . (1 - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{9}$ + .... + $\frac{1}{2001}$ - $\frac{1}{2005}$)
A = -2 . (1 - $\frac{1}{2005}$)
A = -2 . $\frac{2004}{2005}$
A = $\frac{-4008}{2005}$.
B = -$\frac{-3}{2.5}$ - $\frac{3}{5.8}$ - ... - $\frac{3}{2003.2006}$
B = $\frac{3}{2.5}$ - $\frac{3}{5.8}$ - ... - $\frac{3}{2003.2006}$
B = -($\frac{3}{2.5}$ + $\frac{3}{5.8}$ + ... + $\frac{3}{2003.2006}$)
B = -($\frac{1}{2}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{8}$ + ... + $\frac{1}{2003}$ - $\frac{1}{2006}$)
B = - ($\frac{1}{2}$ - $\frac{1}{2006}$)
B = $\frac{-501}{2003}$.
Vì $\frac{-4008}{2005}$ > -1
Mà $\frac{-501}{2003}$ < -1
=> $\frac{-4008}{2005}$ > $\frac{-501}{2003}$
=> A > B.
