Question

已知2x-3y=0,3x-2y-6z=0,且xyz不等于0,求 x^2+y^2+z ———————— 2x^2+y^2-

Answer 1

由2x-3y+z=0 ==>2x-3y=-z ①3x-2y=6z ②

把z看成常数 则由②×2-①×3可得 5y=15z ==>y=3z

代入②可得3x-2×3z=6z ==>x=4z

所以(x2+y2+z2)/(xy+yz+xz)=[(4z)2+(3z)2+z2]/(4z*3z+3z*z+4z*z)=(16z2+9z2+z2)/(12z2+3z2+4z2)=26z2/16z2=13/8