(1+2xy)^2-(x^2+y^2)^2速求
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(1+2xy)^2-(x^2+y^2)^2速求
因式分解
因式分解
原式=[(1+2xy)+(x²+y² )][(1+2xy)-(x² +y² )]
=-(x²+y² +2xy+1)(x² +y² -2xy-1)
=-(x²+y² +2xy+1)[(x-y)² -1)]
=-(x²+y² +2xy+1)[(x-y)+1][(x-y)-1]
=-(x-y+1)(x-y-1)(x²+y² +2xy+1)
=-(x²+y² +2xy+1)(x² +y² -2xy-1)
=-(x²+y² +2xy+1)[(x-y)² -1)]
=-(x²+y² +2xy+1)[(x-y)+1][(x-y)-1]
=-(x-y+1)(x-y-1)(x²+y² +2xy+1)
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