已知x是一元二次方程x²+3x-1=0的实数根,那么代数x-3/(3x²-6x)

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= (x-3) / [3x(x-2)] / [ (x^2 - 4 - 5)/(x-2) ]
= (x-3) / [3x(x-2) * (x+3)(x-3)/(x-2)]
= 1/ [3x(x+3)]
= (1/ 3 ) * 1/ (x^2 + 3x) (代入一元二次方程)
= 1/3.这是正确的做法,
======以下答案可供参考======
供参考答案1:
1/3供参考答案2:
自己没写清楚，分子分母都看不懂。
供参考答案3:
原式= (x-3) / [3x(x-2)] / [ (x^2 - 4 - 5)/(x-2) ]
= (x-3) / [3x(x-2) * (x+3)(x-3)/(x-2)]
= 1/ [3x(x+3)]
= (1/ 3 ) * 1/ (x^2 + 3x) (代入一元二次方程)
= 1/3.