`(x - 1)/2011 + (x - 2)/2010 - (x - 3)/2009 = (x - 4)/2008`
`⇒ ((x - 2012) + 2011)/2011 + ((x - 2012) + 2010)/2010 - ((x - 2012) + 2009)/2010 - ((x - 2012) + 2008)/2008 = 0`
`⇒ (x - 2012)/2011 + 1 + (x - 2012)/2010 + 1 - (x - 2012)/2009 - 1 - (x - 2012)/2008 - 1 = 0`
`⇒ (x - 2012). (1/2011 + 1/2010 - 1/2009 - 1/2008) = 0`
Ta thấy: `1/2011 + 1/2010 - 1/2009 - 1/2008 \neq 0`
`⇒ x - 2012 = 0 ⇔ x = 2012`
