已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2的值

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已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2的值已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+1

已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2的值
已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2的值

已知(x-3)^2+(y+1)^2+z^2=0,求x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2的值
(x-3)^2+(y+1)^2+z^2=0
则
x-3=0
y+1=0
z=0
x=3
y=-1
z=0
x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2
=(x^2-5x^2-2x^2)-(2xy-3xy)+(12xz-8xz)-z^2
=-6x^2+xy+4xz-z^2
=-6*3^2+3*(-1)+4*3*0-0^2
=-54-3+0-0
=-57

由(x-3)^2+(y+1)^2+z^2=0可得到x=3.y=-1,z=0.
x^2-2xy-5x^2+12xz+3xy-z^2-8xz-2x^2
=-6x^2+xy+4xz-z^2=-6×3^2+3×(-1)=-57

因为(x-3)^2>=0，(y+1)^2>=0，z^2>=0
由已知得 x=3,y=-1,z=0
所以原式=-57

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