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已知x-y=1,求代数式x4-xy3-x3y-3x2y+3xy2+y4
人气:368 ℃ 时间:2019-08-17 14:44:21
解答
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(x-y)[(x-y)(x2+xy+y2)-3xy]=1×[1×(x2+xy+y2)-3xy]=x2-2xy+y2=(x-y)2...
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