giải phương trình bậc nhất x+1/2007+x+2/2006=x+2001/7+x+2002/6

$\dfrac{x+1}{2007} + \dfrac{x+2}{2006} = \dfrac{x+2001}{7} + \dfrac{x+2002}{6}$

$⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 2 = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 2$

$⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 1 + 1  = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 1 + 1$

$⇔ (\dfrac{x+1}{[2007}+1) + (\dfrac{x+2}{2006}+1) = (\dfrac{x+2001}{7}+1) + (\dfrac{x+2002}{6}+1)$

$⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} = \dfrac{x+2008}{7} + \dfrac{x+2008}{6}$

$⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} – \dfrac{x+2008}{7} – \dfrac{x+2008}{6} = 0$ 

$⇔ ( x + 2008 ) + ( \dfrac{1}{2007} + \dfrac{1}{2006} – \dfrac{1}{7} – \dfrac{1}{6} ) = 0 $

$⇔ x + 2008 = 0$ 

$⇔ x = -2008$