x^2+x-3=0,求(3-x^2-x^3)/(x-1)的值
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解:已知变形为:x^2=3-x，两边同乘以x，得:x^3=3x-x^2=3x-(3-x)=4x-3，
所以
(3-x^2-x^3)/(x-1)
=[3-(3-x)-(4x-3)]/(x-1)
=(-3x+3)/(x-1)
=-3(x-1)/(x-1)
=-3
方法二:x^2+x=3
(3-x^2-x^3)/(x-1)
=[3-(x^2+x^3)]/(x-1)
=[3-x(x+x^2)]/(x-1)
=[3-x*3]/(x-1)
=-3(x-1)/(x-1)
=-3
所以
(3-x^2-x^3)/(x-1)
=[3-(3-x)-(4x-3)]/(x-1)
=(-3x+3)/(x-1)
=-3(x-1)/(x-1)
=-3
方法二:x^2+x=3
(3-x^2-x^3)/(x-1)
=[3-(x^2+x^3)]/(x-1)
=[3-x(x+x^2)]/(x-1)
=[3-x*3]/(x-1)
=-3(x-1)/(x-1)
=-3
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