若x,y,z满足3x+7x+z=1,4x+10y+z=2006,求分式2005x+2005y+2005z/x+3y的值
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若x,y,z满足3x+7x+z=1,4x+10y+z=2006,求分式2005x+2005y+2005z/x+3y的值
3x+7x+z=1 ①
4x+10y+z=2006 ②
②-①得
x+3y=2005 ③
①乘以3得
9x+21y+3z=3 ④
②乘以2得
8x+20y+2z=4012 ⑤
④-⑤得
x+y+z=-4009
所以所求式子
=2005(x+y+z)/(x+3y)
=2005*(-4009)/2005
=-4009
用4x+10y+z=2006减去3x+7y+z=1
得x+3y=2005
用3x+7y+z=1乘以3得9x+21y+3z=3-----P
用4x+10y+z=2006乘以2得8x+20y+2z=4012-----Q
用方程P减去方程Q得x+y+z=-4009
所以所求式子为(2005*4009)/(-2005)=-4009
明教为您解答,
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用4x+10y+z=2006减去3x+7y+z=1
得x+3y=2005
用3x+7y+z=1乘以3得9x+21y+3z=3-----P
用4x+10y+z=2006乘以2得8x+20y+2z=4012-----Q
用方程P减去方程Q得x+y+z=-4009
所以所求式子为(2005*4009)/(-2005)=-4009
明教为您解答,
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