若x,y,z满足3x+7y+z=1,4X+10y+z=2008,求分式(2007x+2007y+2007z)/(x+3y)的值

来源：百度知道编辑：UC知道时间：2024/06/06 09:28:36

1/(x^2-2x+4)* (x^2-4/x+8) + 1/(x+8)
~~~~~~~~~~~~~~

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

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)
=

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