已知x/3=y/4=z/2≠0则2x^-2y^2+5z^2/xy+yz+zx

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已知x/3=y/4=z/2≠0则2x^-2y^2+5z^2/xy+yz+zx
已知x/3=y/4=z/2≠0则2x^-2y^2+5z^2/xy+yz+zx

已知x/3=y/4=z/2≠0则2x^-2y^2+5z^2/xy+yz+zx
设 x/3=y/4=z/2=t ,
则 x=3t ,y=4t ,z=2t ,
代入可得 (2x^2-2y^2+5z^2)/(xy+yz+zx)=(18t^2-32t^2+20t^2)/(12t^2+8t^2+6t^2)
=(18-32+20)/(12+8+6)
=3/13.

x/3=y/4
x=3y/4

y/4=z/2
z=y/2

2x^-2y^2+5z^2/xy+yz+zx
=[2(3y/4)²-2y²+5(y/2)²]/[3y/4*y+y*y/2+3y/4*y/2]
=(9/8y²-2y²+5/4y²)/(3/4y²+1/2y²+3/2y²)
=3/8y² / 11/4y²
=3/22

令x/3=y/4=z/2=k
则x=3k, y=4k ,z=2k
2x^2-2y^2+5z^2/xy+yz+zx
=(18k^2-32k^2+20k^2)/(12k^2+8k^2+6k^2)
=3/13

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