(x+y)(x+y+2xy)+(xy+1)(xy-1)分解因式

问题描述:

(x+y)(x+y+2xy)+(xy+1)(xy-1)分解因式

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)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)