`p) ( x + 1 )/2012 + ( x + 2 )/2011 + ( x + 3 )/2010 = ( x + 10 )/2003 + ( x + 11 )/2002 + ( x + 12 )/2001`
`⇔ ( ( x + 1 )/2012 + 1 ) + ( ( x + 2 )/2011 + 1 ) + ( ( x + 3 )/2010 + 1 ) = ( ( x + 10 )/2003 + 1 ) + ( ( x + 11 )/2002 + 1 ) + ( ( x + 12 )/2001 + 1 )`
`⇔ ( x + 2013 )/2012 + ( x + 2013 )/2011 + ( x + 2013 )/2010 = ( x + 2013 )/2003 + ( x + 2013 )/2002 + ( x + 2013 )/2001`
`⇔ ( x + 2013 )/2012 + ( x + 2013 )/2011 + ( x + 2013 )/2010 - ( x + 2013 )/2003 - ( x + 2013 )/2002 - ( x + 2013 )/2001 = 0`
`⇔ ( x + 2013 ) . ( 1/2012 + 1/2011 + 1/2010 - 1/2003 - 1/2002 - 1/2001 ) = 0`
`⇔ x + 2013 = 0`
`⇔ x = 0 - 2013`
`⇔ x = - 2013`
Vậy `, x = - 2013 .`
