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分解因式（x+y-2xy)(x+y-2)+(xy-1)２同上

2006-07-08 21:27:431***
同上分解因式（x+y-2xy)(x+y-2)+(xy-1)２同上：别怕烦，全解出来，则原式变为：（用的方法是主元法，把X看成主元，把Y看成常数） （x+y-2xy?

最佳回答

  • 别怕烦，全解出来，则原式变为：（用的方法是主元法，把X看成主元，把Y看成常数） （x+y-2xy)(x+y-2)+(xy-1)２=(y2-2y+1)x2-2(y2-2y+1)x+y2-2y+1 =(y-1)2(x-1)2
    2006-07-08 21:51:48
  • (x+y-2xy)(x+y-2)+(xy-1)^2 =(x+y)^2-2xy(x+y)-2(x+y)+4xy +(xy)^2-2xy+1 =[(x+y)^2-2xy(x+y)+(xy)^2]-2(x+y-xy)+1 =(x+y-xy)^2-2(x+y-xy)+1 =[(x+y-xy)-1]^2 =(-xy+x+y-1)^2 =[-x(y-1)+(y-1)]^2 =[(y-1)(1-x)]^2 =(x-1)^2*(y-1)^2.
    2006-07-09 07:25:03
  • 为清楚起见，令x+y=m，xy=n 则原式=(m-2n)(m-2)+(n-1)^2 =m^2-2m-2mn+4n+n^2-2n+1 =(m-n)^2-2m+2n+1 =(m-n)^2-2(m-n)+1 =(m-n-1)^2 =(x+y-xy-1)^2 =(xy-x-y+1)^2 =(x-1)^2(y-1)^2
    2006-07-08 22:05:37
  • 原式=x^2+y^2-2x-2y-2x^2y-2xy^2+4xy+x^2y^2+1 =(x^2+y^2+2xy)-(2x+2y+2x^2y+2xy^2)+(x^2y^2+2xy+1) =(x+y)^2-2(x+y)(xy+1)+(xy+1)^2 =(x+y-xy-1)^2 =[(x-1)+y(1-x)]^2 =(x-1)^2(y-1)^2
    2006-07-08 21:55:22
  • 晕倒,这种题目我几年前做起来就像是吃个大饼那么简单,怎么现在看起来像天书一样
    2006-07-08 21:33:18
    • 很赞哦！ (257)

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