已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.
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![已知x/a-b=y/b-c=z/c-a.求证x+y+z=0.](/uploads/image/z/10764072-0-2.jpg?t=%E5%B7%B2%E7%9F%A5x%2Fa-b%3Dy%2Fb-c%3Dz%2Fc-a.%E6%B1%82%E8%AF%81x%2By%2Bz%3D0.)
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