题目内容

当x-y=1时,求以下代数式的值:

x4-xy3-x3y-3x2y+3xy2+y4

答案:
解析:

  ∵x-y=1,

  ∴x4-xy3-x3y-3x2y+3xy2+y4

  =(x4-x3y)-(xy3-y4)-3xy(x-y)

  =x3(x-y)-y3(x-y)-3xy(x-y)

  =(x-y)(x3-y3-3xy)

  =x3-y3-3xy

  =(x-y)(x2+xy+y2)-3xy

  =x2+xy+y2-3xy

  =x2-2xy+y2

  =(x-y)2

  =1.


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