(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求:xy+yz+zx/x^2+y^2+z^2

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(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求:xy+yz+zx/x^2+y^2+z^2

(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求:xy+yz+zx/x^2+y^2+z^2
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求:xy+yz+zx/x^2+y^2+z^2

(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求:xy+yz+zx/x^2+y^2+z^2
平方和绝对值都大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个式子都等于0
所以x-3y+z=0 (1)
5x-4y+z=0 (2)
(1)-(1)
4x-y=0
y=4x
(2)-(1)*5
5x-4y+z-5x+15y-5z=0
11y-4z=0
z=11y/4,y=4x
所以z=11x
代入xy+yz+zx/x^2+y^2+z^2
=(4x^2+44x^2+11x^2)/(x^2+16x^2+121x^2)
=59x^2/138x^2
=59/138

(x-3y+z)^2和/ 5x-4y+z/都非负,且相加为0
所以x-3y+z=5x-4y+z=0
于是y=4x ,z=11x
所以xy+yz+zx/x^2+y^2+z^2=
(4+44+11)x^2/(1+16+121)x^2=59/138