将x=1代入原方程,有:a×1+b=c,即:a+b=c
由此得:(a+b)/c=1
又得:a+b-c=0
[(a+b)×(-1/c)]^2005
=-[(a+b)/c]^2005
=-(1)^2005
=-1
即:a+b-c=0,[(a+b)×(-1/c)]^2005=-1
将x=1代入原方程,有:a×1+b=c,即:a+b=c
由此得:(a+b)/c=1
又得:a+b-c=0
[(a+b)×(-1/c)]^2005
=-[(a+b)/c]^2005
=-(1)^2005
=-1
即:a+b-c=0,[(a+b)×(-1/c)]^2005=-1