Đáp án:
Do `x,y,zne0` nên:
`{xy}/{ay+bx}={yz}/{bz+cy}={zx}/{cx+az}=>{zxy}/{ayz+bxz}={xyz}/{bzx+cyx}={yzx}/{cxy+azy}`
Suy ra `ayz+bxz=bzx+cyx=cxy+azy=>az=cx,bx=ay`
Do đó
`x/a=z/c,x/a=y/b=>x/a=y/b=z/c=t=>x=at,y=bt,z=ct,tne0`
Ta có `{xy}/{ay+bx}={x^2+y^2+z^2}/{a^2+b^2+c^2}=>{at.bt}/{abt+bat}={a^2t^2+b^2t^2+c^2t^2}/{a^2+b^2+c^2}`
Suy ra `t/2=t^2=>t=1/2(`do `tne0)`
Vậy `x=a/2,y=b/2,z=c/2`