令a+1/b=b+1/c=c+1/a=kab+1=bk,bc+1=ck,--------所以bc=kc-1ca+1=ak.由ab+1=bk,abc+c=kbc=k(kc-1)abc-k=(k^2-1)c,同样可以算出,abc-k=(k^2-1)b,abc-k=(k^2-1)a,abc-k=(k^2-1)a=(k^2-1)b=(k^2-1)c因为a,b,c互不相...