x-y=1,求x²-y²-2y

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x-y=1,求x²-y²-2yx-y=1,求x²-y²-2yx-y=1,求x²-y²-2yx-y=1所以原式=(x+y)(x-y)-2y=(

x-y=1,求x²-y²-2y
x-y=1,求x²-y²-2y

x-y=1,求x²-y²-2y
x-y=1
所以原式=(x+y)(x-y)-2y
=(x+y)*1-2y
=x+y-2y
=x-y
=1

x=y+1 替换为 (y+1)² -y²-2y
y²+2y+1-y²-2y = 1