Đáp án:
$e) \dfrac{2000-x}{4} + \dfrac{2001-x}{3} = \dfrac{2-x}{2002} + \dfrac{1-x}{2003}$
$⇔\dfrac{2000-x}{4} + \dfrac{2001-x}{3} - \dfrac{2-x}{2002} - \dfrac{1-x}{2003} =0$
$ ⇔ \dfrac{2000-x}{4} + 1 + \dfrac{2001-x}{3} + 1- \dfrac{2-x}{2002} + 1 - \dfrac{1-x}{2003} + 1 =0$
$⇔ \dfrac{2004-x}{4} + \dfrac{2004 -x}{3} - \dfrac{2004-x}{2002} - \dfrac{2004-x}{2003}= 0$
$⇒ (2004-x)(\dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{2002} -\dfrac{1}{2003})=0$
Vì $\dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{2002} -\dfrac{1}{2003} \neq 0 $
$⇔ 2004 - x = 0$
$⇔ -x = -2004$
$⇔ x = 2004$
$\text{Vậy phương trình có tập nghiệm S={2004}}$
