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:
解析:
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∵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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