若x^2+x-3=0,则分式3-x^2-x^3/x-1=?
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若x^2+x-3=0,则分式3-x^2-x^3/x-1=?
RT
RT
![若x^2+x-3=0,则分式3-x^2-x^3/x-1=?](/uploads/image/z/19384283-11-3.jpg?t=%E8%8B%A5x%5E2%2Bx-3%3D0%2C%E5%88%99%E5%88%86%E5%BC%8F3-x%5E2-x%5E3%2Fx-1%3D%3F)
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
(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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