若x/3=y/1=z/2,且xy+yz+zx=99,求zx平方+9y平方+9z平方的值?
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若x/3=y/1=z/2,且xy+yz+zx=99,求zx平方+9y平方+9z平方的值?
若x/3=y/1=z/2,且xy+yz+zx=99,求zx平方+9y平方+9z平方的值?
若x/3=y/1=z/2,且xy+yz+zx=99,求zx平方+9y平方+9z平方的值?
令x/3=y/1=z/2=k
x=3k
y=k
z=2k
xy+yz+zx=99
3k*k+k*2k+3k*2k=99
3k^2+2k^2+6k^2=99
11k^2=99
k^2=9
k=±3
x^2=9k^2=9*9=81
y^2=k^2=9
z^2=4k^2=4*9=36
当k=3时,z=6
zx^2+9y^2+9z^2
=6*81+9*9+9*36
=891
当k=3时,z=-6
zx^2+9y^2+9z^2
=-6*81+9*9+9*36
=-81
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