x-2/2006+2/2007+x+2/2008=6

设y=2007 则y-1=2006 y+1=2008
则x/(y-1)+x/(y+1)=6+2/(y-1)-2/y-2/(y+1)
x*[2y/(y-1)(y+1)]=2(3+1/(y-1)-1/y-1/(y+1)]
yx=3(y²-1)+y+1-(y²-1)/y-(y-1)
yx=3y²-3+y+1-y+1/y-y+1
yx=3y²-y-1+1/y
所以x=3y-1-1/y+1/y²
即x=3*2007-1-1/2007+1/2007²
x=6020-1/2007+1/2007²

设x=2y (y-1)/2006+y/2007+(y+1)/2008=3 很显然 y=2007时，(y-x(1+1/2007+1)=6+1/1003-1/1004 x=(6+1/1003-1/1004)/(2+1/

2x=6-2/2008+2/2006-2/2007
x=3-1/2008+1/2006-1/2007