已知x.y.z均为实数x+y+z≠0 a=x/y+z b=y/z+x c=z/x+y那么a/a+1+b/b+1+c/c+

1个回答

  • ∵x/﹙y+z﹚=a,y/﹙z+x﹚=b,z/﹙x+y﹚=c

    ∴﹙y+z﹚/x=1/a,﹙z+x﹚/y=1/b,﹙x+y﹚/z=1/c

    ∴﹙y+z﹚/x+1=1/a+1 ∴﹙y+z+x﹚/x=﹙a+1﹚/a

    ∴x/﹙x+y+z﹚=a/﹙a+1﹚

    同理y/﹙x+y+z﹚=b/﹙b+1﹚

    z/﹙x+y+z﹚=c/﹙c+1﹚

    ∴a/﹙a+1﹚+b/﹙b+1﹚+c/﹙c+1﹚=x/﹙x+y+z﹚+y/﹙x+y+z﹚+z/﹙x+y+z﹚

    =﹙x+y+z﹚/﹙x+y+z﹚

    =1