Question

【2(xy+1)的平方-2(x的平方+y的平方+1)】除以y

Answer 1

[2(xy+1)²-2(x²+y²+1)]/y=[2(x²y²+2xy+1)-2x²-2y²-2]/y
=(2x²y²+4xy-2x²-2y²)/y
=2x²y+4x-2x²/y-2y

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Answer 2

【2(xy+1)的平方-2(x的平方+y的平方+1)】除以y
=[2x^2y^2+4xy+2-2x^2-2y^2-2]/y
=[2x^2y^2+4xy-2x^2-2y^2]/y
=2x^2y+4x-2y-2x^2/y
这个式子化简也很复杂，有问题吧？

Answer 3

【2(xy+1)的平方-2(x的平方+y的平方+1)】除以y
=(2x²y²+4xy+2-2x²-2y²-2)÷y
=(2x²y²+4xy-2x²-2y²)÷y

Answer 4

[2(xy+1)²-2(x²+y²+1)]/y
=[2(x²y²+2xy+1)-2x²-2y²-2]/y
=(2x²y²+4xy-2x²-2y²)/y
=[2x²y²+2(2xy-x²-y²)]/y
=[2x²y²-2(x²-2xy+y²)]/y
=[2x²y²-2(x-y)²]/y
=2[x²y²-(x-y)²]/y
=2(xy+x-y)(xy-x+y)/y