Question

初二的因式分解题目！

(x+y)(x+y+2xy)+(xy+1)(xy-1)
(x+y)(x-y)+4(y-1)
要有详细过程！

Answer 1

(x+y)(x+y+2xy)+(xy+1)(xy-1)
=(x+y)(x+y)+2xy(x+y)+(xy+1)(xy-1) 乘法分配律
2 2 2 2 2 2
=x +2xy+y +2x y+2xy +x y -1

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

Answer 2

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

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