已知xy=8满足s^2y-8xy^2-x+y=56,求x^2+y^2的值（答案为80）

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已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值（答案为80）
1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值（答案为3）
2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,；若不能,请简要说明理由．

x^2y-8xy^2-x+y=56
xy(x-y)-(x-y)=56
xy=8
所以8(x-y)-(x-y)=56
7(x-y)=56
x-y=8
两边平方
x^2-2xy+y^2=64
x^2+y^2=64+2xy=64+2*8=80
x-3=y-2
所以x-y=1
x-3=z-1
所以x-z=2
y-2=z-1
所以y-z=1
x^2+y^2+z^2-xy-yz-zx
=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2
=[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)]/2
=[(x-y)^2+(y-z)^2+(z-x)^2]/2
=(1^2+1^2+2^2)/2
=3
x^2+y^2-x+6y+41/4
=(x^2-x+1/4)+(y^2+6y+9)+1
=(x-1/2)^2+(y+3)^2+1
完全平方大于等于0
所以(x-1/2)^2+(y+3)^2>=0
所以(x-1/2)^2+(y+3)^2+1>0
所以是正数

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