Các bất phương trình ẩn x sau nha !!!

Đáp án:

 $\dfrac{x+2004}{2005} + \dfrac{x+2005}{2006} < \dfrac{x+2006}{2007}+ \dfrac{x+2007}{2008}$

$<=> \dfrac{x+2004}{2005} -1 + \dfrac{x+2005}{2006}-1 < \dfrac{x+2006}{2007}-1 + \dfrac{x+2007}{2008} -1 $

$<=> \dfrac{x+2004-2005}{2005} + \dfrac{x+2005-2006}{2006} < \dfrac{x+2006-2007}{2007} + \dfrac{x+2007 -2008}{2008}$

$<=>\dfrac{x-1}{2005} + \dfrac{x-1}{2006} < \dfrac{x-1}{2007} + \dfrac{x-1}{2008}$

$<=> \dfrac{x-1}{2005} + \dfrac{x-1}{2006} - \dfrac{x-1}{2007} - \dfrac{x-1}{2008} < 0 $

$<=> (x-1)( \dfrac{1}{2005} + \dfrac{1}{2006} - \dfrac{1}{2007} - \dfrac{1}{2008} ) <0$

Vì $\dfrac{1}{2005} + \dfrac{1}{2006} - \dfrac{1}{2007} - \dfrac{1}{2008} \neq 0 $

$=> x-1 <0$

$<=> x <1$

$\text{Vậy bất phương trình có tập nghiệm S = {$x | x <1$ } }$

$ b) \dfrac{x-2}{2002} + \dfrac{x-4}{2000} < \dfrac{x-3}{2001} + \dfrac{x-5}{1999}$

$<=>\dfrac{x-2}{2002} -1  + \dfrac{x-4}{2000} -1  < \dfrac{x-3}{2001} -1  + \dfrac{x-5}{1999} -1 $

$<=> \dfrac{x-2-2002}{2002} + \dfrac{x-4-2000}{2000} < \dfrac{x-3-2001}{2001} + \dfrac{x-5-1999}{1999}$ 

$<=> \dfrac{x-2004}{2002} + \dfrac{x-2004}{2000} < \dfrac{x-2004}{2001} + \dfrac{x-2004}{1999}$

$<=> \dfrac{x-2004}{2002} + \dfrac{x-2004}{2000} - \dfrac{x-2004}{2001} - \dfrac{x-2004}{1999} < 0 $

$<=>(x-2004)( \dfrac{1}{2002} + \dfrac{1}{2002} - \dfrac{1}{2001}- \dfrac{1}{1999}) <0$

Vì $\dfrac{1}{2002} + \dfrac{1}{2002} - \dfrac{1}{2001}- \dfrac{1}{1999} \neq 0 $

$=> x-2004 <0$

$<=> x < 2004$

$\text{Vậy bất phương trình có tập nghiệm S = {$x| x < 2004$} }$