若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=

问题描述:

若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=

令X=3K Y=K Z=2K 代入得:3K^2+2K^2+6K^2=99 所以K=3或-3,故X^2=81 Y^2=9 Z^2=36 所以原式=594