x²-y²-3x-3y

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x²-y²-3x-3yx²-y²-3x-3yx²-y²-3x-3yx²-y²-3x-3y=(x-y)(x+y)-3(x+

x²-y²-3x-3y
x²-y²-3x-3y

x²-y²-3x-3y
x²-y²-3x-3y
=(x-y)(x+y)-3(x+y)
=(x+y)(x-y-3)

问什么?

=(x+y)(x-y)-3(x+y)
=(x+y)(x-y-3)