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(2012·兰州)已知x是一元二次方程x²-2x+1=0的根,求代数式(x-3)/(3x²-6x)÷

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(2012·兰州)已知x是一元二次方程x²-2x+1=0的根,求代数式(x-3)/(3x²-6x)÷[x+2-(5/x-2)]的值.

x²-2x+1=0
(x-1)²=0
x1=x2=1
[(x-3)/(3x²-6x)] ÷ [(x+2)-5/(x-2)]
={(x-3)/[3x(x-2)]} ÷ {[(x+2)(x-2)-5]/(x-2)}
= {(x-3)/[3x(x-2)]} × {(x-2)/(x+3)(x-3)}
=1/[3x(x+3)]
因此:
带入x1=x2=1,则:
[(x-3)/(3x²-6x)] ÷ [(x+2)-5/(x-2)]
=1/[3x(x+3)]
=1/12