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当x-y=1时,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值
人气:166 ℃ 时间:2019-09-23 14:52:04
解答
x^4-xy^3-x^3y-3x^2y+3xy^2+y^4 =(x^4-x^3y)-xy^3+y^4-3x^2y+3xy^2 =x^3(x-y)-y^3(x-y)-3xy(x-y) =x^3-y^3-3xy =(x-y)(x^2+xy+y^2)-3xy =x^2+xy+y^2-3xy =x^2-2xy+y^2 =(x-y)^2 =1
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