Question

（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)÷

（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.

Answer 1

解：
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

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