已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.

已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.
已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.

已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.
令x/a-b=y/b-c=z/c-a=k
则x=k(a-b)
y=k(b-c)
z=k(c-a)
所以x+y+z
=k(a-b+b-c+c-a)
=k*0
=0

x/a-b=y/b-c=z/c-a.=k
a+y+z=(a-b+b-c+c-a)k
=0