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求值域求x^2/(x^2-x+1)值域

2010-07-24 08:19:27圆***
求x^2/(x^2-x+1)值域。求值域求x^2/(x^2-x+1)值域。：设y=x^2/(x^2-x+1),则 y(x^2-x+1)=x^2, (y-1)x^2-yx+y=0, ∵x∈R?

最佳回答

  • 设y=x^2/(x^2-x+1),则 y(x^2-x+1)=x^2, (y-1)x^2-yx+y=0, ∵x∈R, ∴y^2-4y(y-1)>=0, 即y(y-4/3)<=0, ∴0<=y<=4/3,为所求。
    2010-07-24 08:58:55
  • x^2/(x^2-x+1)=1/(1-1/x+1/x^2) x不为0 换元u=1/x，原式=1/（1-u+u^2) u不为0 1-u+u^2值域为[0.75,正无穷)原函数值域（0,4/3] x=0时为0 综上值域[0,4/3]
    2010-07-24 11:10:30
    • 很赞哦！ (122)

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