已知x/(b-c)yz=y/(c-a)xz=z/(a-b)xy,求证：x求证：x^2+y^2+z^2=0

Question

已知x/(b-c)yz=y/(c-a)xz=z/(a-b)xy,求证：x
求证：x^2+y^2+z^2=0

Answer 1

等式x/(b-c)yz=y/(c-a)xz=z/(a-b)xy同时乘xyz,得到
x^2/(b-c)=y^2/(c-a)=z^2/(a-b)
∴ y^2=(c-a)/(b-c)*x^2
z^2=(a-b)/(b-c)*x^2
带入x^2+y^2+z^2=0 得证