a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)

设a/x^2-yz=b/y^2-zx=c/z^2-xy =k因为xyz=0所以x,y,z中至少有一个=0不妨设x=0则a/x^2-yz=b/y^2-zx=c/z^2-xya/-yz=b/y^2=c/z^2=ka=-kyzb=ky^2c=kz^2ax+by+cz=by+cz=ky^3+kz^3=k(y^3+z^3)(a+b+c)(x+y+z)=k(y^2+z^2-yz)...