因式分解(x+y)(x+y+2xy)+(xy+1)(xy-1)已得出答案令x+y=a,xy=b,则原式=a(a+2b)+(b+1)(b-1)=a^2+2ab+b^2-1=(a+b)^2-1=(a+b+1)(a+b-1)故原式=(x+y+xy+1)(x+y+xy-1)=(x+1)(y+1)(x+y+xy-1)故原式=(x+y+xy+1)(x+y+xy-1)=(x+1)(y+1)(x+y+xy-1)这个步骤

问题描述:

因式分解(x+y)(x+y+2xy)+(xy+1)(xy-1)
已得出答案令x+y=a,xy=b,则
原式=a(a+2b)+(b+1)(b-1)
=a^2+2ab+b^2-1
=(a+b)^2-1
=(a+b+1)(a+b-1)
故原式=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
故原式=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)这个步骤

(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
是因为 x+y+xy+1=x(y+1)+(y+1)=(x+1)(y+1)
这是分组分解法