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因式分解:(1)a4-16(2)(x-y)2+4xy

(1)a4-16=(a2+4)(a2-4)=(a2+4)(a+2)(a-2);(2)(x-y)2+4xy=x2-2xy+y2+4xy=x2+2xy+y2=(x+y)2.

x2-2xy+y2+4xy= x2+2xy+y2=(x+y)2

(x-y)2+4xy=x2-2xy+y2+4xy,=x2+2xy+y2,=(x+y)2.故答案为:(x+y)2.

(1-x^2)(1-y^2)+4xy因式分解=1-y^2-x^2+x^2y^2+4xy=[x^2y^2+2xy+1]-[x^2-2xy+y^2]=[xy+1]^2-[x-y]^2=[xy+1+x-y][xy+1-x+y]

x2+y2-x2y2-4xy-1=x^2-2xy+y^2-(x^2y^2+2xy+1)=(x-y)^2-(xy+1)^2=(x-y+xy+1)(x-y-xy-1)

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

(X-Y)(X-Y)+4XY =X^2-2XY+Y^2+4XY =X^2+2XY+Y^2 =(X+Y)^2

因式分解:(1)y(x-y)^2+4xy^2解,得:==y(x^2-2xy+y^2)+4xy^2==x^2y-2xy^2+y^3+4xy^2==x^2y+2xy^2+y^3==y(x^2+2xy+y^2)==y(x+y)^2(2)y-(x-y)^2+4xy^2解,得:==y-[(x-y)^2-4xy^2]==y-[(x-y)+2xy]*[(x-y)+2xy]==y-((好像后面这题有点问题~,问一下题目是对的么?~@)

1.=x^2-2xy+y^2+4xy =x^2+2xy+y^2 =(x+y)^22.=4a^2-12ab+9b^2 =(2a-3b)^2^2为平方

1/2xy+2xy+2=1/2(xy+4xy+4)=1/2[(xy)+4xy+4]=1/2(xy+2)

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