因式分解 （x+y)(x+y+2xy)+x的平方y的平方-1
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解：
（x+y)(x+y+2xy)+x的平方y的平方-1
=[(x+y)²+2(x+y)xy+x²y²]-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
（x+y)(x+y+2xy)+x的平方y的平方-1
=[(x+y)²+2(x+y)xy+x²y²]-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
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原式=(x+y)²+2xy(x+y)+x²y²-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=[x(y+1)+(y+1)](x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=[x(y+1)+(y+1)](x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
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(x+y)(x+y+2xy)+x²y²-1
=(x+y)[(x+y)+2xy]+x²y²-1
=(x+y)²+2xy(x+y)+x²y²-1
=(x+y+xy)²-1
=(x+y+xy-1)(x+y+xy+1)
=(x+y+xy-1)(x+1)(y+1)
=(x+y)[(x+y)+2xy]+x²y²-1
=(x+y)²+2xy(x+y)+x²y²-1
=(x+y+xy)²-1
=(x+y+xy-1)(x+y+xy+1)
=(x+y+xy-1)(x+1)(y+1)
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