已知,x/3=y/4=z/5,求x^2+y^2+z^2/xy+yz+xz的值

Question

已知,x/3=y/4=z/5,求x^2+y^2+z^2/xy+yz+xz的值

Answer 1

设x/2=y/3=z/4=a 则:x=2a;y=3a;z=4a 代入得：(xy+yz+zx)/(x^2+y^2+z^2) =(6a^2+12a^2+8a^2)/(4a^2+9a^2+16a^2) =26