因式分解(x+y)(x+y+2xy)+(xy+1)(xy-1)
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因式分解(x+y)(x+y+2xy)+(xy+1)(xy-1)
因式分解(x+y)(x+y+2xy)+(xy+1)(xy-1)
因式分解(x+y)(x+y+2xy)+(xy+1)(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)
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