几道初一因式分解题1.（x-y)²+4xy2.(x-1)(x-2)-2(2-x)²3.(a²+ab+b²)²-9a²b²

1.
（x-y)^2+4xy
=x^2+y^2-2xy+4xy
=x^2+y^2+2xy
=(x+y)^2
2.
(x-1)(x-2)-2(2-x)^2
=(x-1)(x-2)+2(x-2)^2
=(x-2)[(x-1)+2(x-2)]
=(x-2)(x-1+2x-4)
=(x-2)(3x-5)
3.
(a^2+ab+b^2)^2-9a^2b^2
=(a^2+ab+b^2)^2-(3ab)^2
=[(a^2+ab+b^2)-(3ab)][(a^2+ab+b^2)+(3ab)]
=(a^2+b^2-2ab)(a^2+b^2+4ab)
=(a-b)^2(a^2+b^2+4ab)

1.=x2-2xy+y2+4xy=(x+y)2
2.=x2-3x+2-8+8x-2x2=-(x2+5x+6)=-(x+2)(x+3)
3.=a4+b4+a2b2+2a3b+2ab3+3a2b2-9a2b2=a4+b4+2a3b+2ab3-6a2b2
=( a4+b4-2 a2b2)+2ab(a2+b2-2ab)=(a2-b2)2+2ab(a-b)2=[(a+b)(a-b)]2+2ab(a-b)2=(a-b)2[(a+b)2+2ab]

1、原式=x²-2xy+y²+4xy=x²+2xy+y²=(x+y)²2、原式=(x-1)(x-2)+(4-2x)(x-2)=(x-1+4-2x)(x-2)=(3-x)(x-2)3、原式=(a²+ab+b²)²-(3ab)²=(a²+ab+b²+3ab)(a²+...

1.（x-y)²+4xy
=x²-2xy+y²+4xy
=x²+2xy+y²
=(x+y)²
2.(x-1)(x-2)-2(2-x)²
=(x-1)(x-2)-2(x-2)²
=(x-2)(x-1-2)
=(x-2)(x-3)
3.(a²+ab+b²)²-9a²b²
=(a²+ab+b²)²-(3ab)²
=(a²+ab+b²+3ab)(a²+ab+b²-3ab)
=(a²+4ab+b²)(a²-2ab+b²)
=(a²+4ab+b²)(a-b)²