Question

x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方

求（yz+1)(zx+1)(xy+1)/(x平方+1)(y平方+1)(z平方+1)
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过程详细

Answer 1

设a=x-y,b=y-z,-a-b=z-x
(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方
b^2+a^2+(-a-b)^2=(-a-b-a)^2+(a-b)^2+(b+a+b)^2
a^2+b^2+a^2+2ab+b^2=4a^2+4ab+b^2+a^2-2ab+b^2+4b^2+4ab+a^2
2a^2+2b^2+2ab=6a^2+6b^2+6ab
4a^2+4ab+4b^2=0
a^2+ab+b^2=0
(x-y)^2+(x-y)(y-z)+(y-z)^2=0
x^2-2xy+y^2+xy-xz-y^2+yz+y^2-2yz+z^2=0
x^2+y^2+z^2-xy-yz-xz=0
(x-y)^2+(y-z)^2+(z-x)^2=0

所以x=y=z
（yz+1)(zx+1)(xy+1)/(x平方+1)(y平方+1)(z平方+1)=1

Answer 2

(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2

2xx+2yy+2zz-2xy-2xz-2zy=6xx+6yy+6zz-6xy-6yz-6zx
xx+yy+zz=xy+yz+zx
(x-y)^2+(y-z)^2+(z-x)^2=0

x=y=z

(yz+1)(zx+1)(xy+1)/(x^2+1)(y^2+1)(z^2+1) =1