已知(x-y)(x-y+16)+64=0,且x²y²-8xy+16=0,求x²+y²的值,
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已知(x-y)(x-y+16)+64=0,且x²y²-8xy+16=0,求x²+y²的值,
即(x-y)²+16(x-y)+64=0
(x-y+8)²=0
x-y=-8
两边平方
x²-2xy+y²=64
且(xy-4)²=0
xy=4
所以x²+y²
=64+2xy
=72
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