x+y-1的绝对值+（x-y+3)²=0,则x²-y²=

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x+y-1的绝对值+（x-y+3)²=0,则x²-y²=
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x+y-1的绝对值+（x-y+3)²=0,则x²-y²=
x+y-1的绝对值+（x-y+3)²=0,则x²-y²=

x+y-1的绝对值+（x-y+3)²=0,则x²-y²=
x+y-1的绝对值+（x-y+3)²=0
所以
x+y-1=0
x-y+3=0
所以
x+y=1
x-y=-3
所以x²-y²
=(x+y)(x-y)
=1×(-3)
=-3
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