已知x+4y-3z=0;4x-5y+2z=0,求X2+y2+z2/xy+yz+zx的值
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由x+4y-3z=0;4x-5y+2z=0,得兄弯
x+4y=3z;(1)
4x-5y=-2z,(2)
(1)*5+(2)*4,得,
x=z/3,
y=(2/3)z,
代稿茄人到
(X^2+y^2+z^2)/(xy+yz+zx)
=(z^2/9+4z^2/9+z^2)/(2z^2/9+2z^2/3+z^2/3)
=(1/9+4/键尘察9+1)/(2/9+2/3+1/3)
=(14/9)/(11/9)
=14/11
x+4y=3z;(1)
4x-5y=-2z,(2)
(1)*5+(2)*4,得,
x=z/3,
y=(2/3)z,
代稿茄人到
(X^2+y^2+z^2)/(xy+yz+zx)
=(z^2/9+4z^2/9+z^2)/(2z^2/9+2z^2/3+z^2/3)
=(1/9+4/键尘察9+1)/(2/9+2/3+1/3)
=(14/9)/(11/9)
=14/11
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