把多项式(xy-1)²+(x+y-2)(x+y-2xy)分解因式

khaki8 1年前 已收到2个回答 举报

僵尸婆仔 春芽

共回答了13个问题采纳率:92.3% 举报

(xy-1)²+(x+y-2)(x+y-2xy)
=x²y²-2xy+1+(x+y)²-2xy(x+y)-2(x+y)+4xy
=x²y²+2xy+1+(x+y)²-2xy(x+y)-2(x+y)
=(xy+1)²-2(xy+1)(x+y)+(x+y)²
=(xy+1-x-y)²
=[x(y-1)-(y-1)]²
=(x-1)²(y-1)²

1年前

5

ng0dfslkfdslkng 花朵

共回答了3867个问题 举报

(xy-1)^2+(x+y-2)(x+y-2xy)
=x^2y^2-2xy+1+(x+y)^2-2(x+y)-2xy(x+y)+4xy
=x^2y^2+2xy+1+(x+y)^2-2(x+y)(xy+1)
=(xy+1)^2-2(x+y)(xy+1)+(x+y)^2
=(xy+1-x-y)^2
=(x-1)^2(y-1)^2

1年前

1
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