当x-y=1时,那么x4-xy3-x3y-3x2y+3xy2+y4的值是( )A.-1B.0C.1D.2
题型:单选题难度:一般来源:不详
当x-y=1时,那么x4-xy3-x3y-3x2y+3xy2+y4的值是( ) |
答案
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. |
举一反三
将下列多项式因式分 (1)m(m-2)-3(2-m) (2)a4-2a2+1 (3)y4-16x4 (4)a3b2-a5. |
在(x+y)(x-y)=x2-y2中,从左向右的变形是______,从右向左的变形是______ |
下列由左到右的变形,属于因式的分解是( )A.(x+2)(x-2)=x2-4 | B.x2-4=(x+2)(x-2) | C.x2-4+3x=(x+2)(x-2)+3x | D.6x2y=2x2•3y |
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