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解由（1-x²）（1-y²）+4xy
=x^2y^2-x^2-y^2+1+4xy
=x^2y^2+2xy+1-(x^2-2xy+y^2)
=(xy+1)^2-(x-y)^2
=[(xy+1)+(x-y)][(xy+1)-(x-y)]
=(xy+x-y+1)(xy-x+y+1)
=x^2y^2-x^2-y^2+1+4xy
=x^2y^2+2xy+1-(x^2-2xy+y^2)
=(xy+1)^2-(x-y)^2
=[(xy+1)+(x-y)][(xy+1)-(x-y)]
=(xy+x-y+1)(xy-x+y+1)
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1-y²-x²+x²y²+4XY
=1-X²Y²+2XY-(X²-2XY+Y²)
=1-XY(XY+2)-(X-Y)²
=(1+X-Y)(1-X+Y)-XY(XY+2)
=1-X²Y²+2XY-(X²-2XY+Y²)
=1-XY(XY+2)-(X-Y)²
=(1+X-Y)(1-X+Y)-XY(XY+2)
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（1-x²）（1-y²）+4xy
=x²y²+2xy+1-x²-y²+2xy
=(xy+1)^2-(x-y)^2
=(xy+1+x-y)(xy+1-x+y)
=x²y²+2xy+1-x²-y²+2xy
=(xy+1)^2-(x-y)^2
=(xy+1+x-y)(xy+1-x+y)
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