已知: x/a+y/b+z/c=1,且a/x+b/y+c/z=0.
求證:(x/a)*(x/a)+(y/b)*(y/b)+(z/c)*(z/c)=0
謝謝....




回應

2005-04-13 14:36
(x/a)*(x/a)+(y/b)*(y/b)+(z/c)*(z/c)
=((bcx)2+(acy)2+(abz)2)/(abc)2
(x/a+y/b+z/c=1-->bcx+acy+abz=abc)
=((abc)2-2*(abc*cxy+abc*ayz+abc*bxz))/(abc)2
=1-2*(cxy+ayz+bxz)/abc
(a/x+b/y+c/z=0-->ayz+bxz+cxy=0)
=1
??????