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把多项式(xy-1)^2+(x+y-2)(x+y-2)(x+y-2xy)分解因式,再求值(x=6,y=11)
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把多项式(xy-1)^2+(x+y-2)(x+y-2)(x+y-2xy)分解因式,再求值(x=6,y=11)
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答案和解析
(XY-1)2 +(x + y的-2)(x + y的-2XY)
= X 2 Y 2-2XY +1 +(+ y)的2-2XY(+ y)的-2 (X + Y)+4 XY
= X 2 Y 2 +2 +1 XY +(X + Y)2-2XY(X + Y)-2(X + Y)
=(XY 1)2 -2板(xy +1)(X + Y)+(+ y)的2
=板(xy +1 xy)的2
= [×(γ-1) - ( γ-1)] 2
=(x-1的)2(γ-1)的2
= X 2 Y 2-2XY +1 +(+ y)的2-2XY(+ y)的-2 (X + Y)+4 XY
= X 2 Y 2 +2 +1 XY +(X + Y)2-2XY(X + Y)-2(X + Y)
=(XY 1)2 -2板(xy +1)(X + Y)+(+ y)的2
=板(xy +1 xy)的2
= [×(γ-1) - ( γ-1)] 2
=(x-1的)2(γ-1)的2
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