若x,y,z满足3x+7y+z=1,4X+10y+z=2008,求分式(2007x+2007y+2007z)/(x+3y)的值1/(x^2-2x+4)* (x^2-4/x+8) + 1/(x+8)
问题描述：
若x,y,z满足3x+7y+z=1,4X+10y+z=2008,求分式(2007x+2007y+2007z)/(x+3y)的值
1/(x^2-2x+4)* (x^2-4/x+8) + 1/(x+8)
答
3x+7y+z=1 (1)
4X+10y+z=2008 (2)
(2)-(1) 得 x+3y=2007
(1)式可化为 x+y+z+2x+6y=1
即 (x+y+z)+2(x+3y)=1
将x+3y=2007代入得 (x+y+z)=1-2*2007=-4013
(2007x+2007y+2007z)/(x+3y)=2007(x+y+z)/(x+3y)
2007(x+y+z)/2007=x+y+z=-4013
1/(x^2-2x+4)* (x^2-4/x+8) + 1/(x+8)
=x/(x²-2x+4)(x³+8x-4)+1/(x+8)
=
答
3x+7y+z=1.1
4X+10y+z=2008.2
2-1:x+3y=2007
所以(2007x+2007y+2007z)/(x+3y)=x+y+z
4X+10y+z=2008中x+3y=2007
所以x+y+z+3(x+3y)=2008
x+y+z=2008-3*2007=-4013
(2007x+2007y+2007z)/(x+3y)=-4013