题目内容
当x-y=1时,那么x4-xy3-x3y-3x2y+3xy2+y4的值是( )
A.-1 | B.0 | C.1 | D.2 |
x4-xy3-x3y-3x2y+3xy2+y4
=(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]
把x-y=1代入得,
原式=1×[1×(x2+xy+y2)-3xy]
=x2-2xy+y2=(x-y)2
∵x-y=1,
∴原式=1.
故选C.
=(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]
把x-y=1代入得,
原式=1×[1×(x2+xy+y2)-3xy]
=x2-2xy+y2=(x-y)2
∵x-y=1,
∴原式=1.
故选C.
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