已知a,b,c互不相等,且1/(x+a)+1/(y+a)+1/(z+a)=1/a,1/(x+b)+1/(y+b)+1/(z+b)=1/b,1/(x+c)+1/(y+c)+1/(z+c)求证:1/a+1/b+1/c=0

Question

已知a,b,c互不相等,且1/(x+a)+1/(y+a)+1/(z+a)=1/a,1/(x+b)+1/(y+b)+1/(z+b)=1/b,1/(x+c)+1/(y+c)+1/(z+c)
求证:1/a+1/b+1/c=0

Answer 1

构造以m为未知数的方程1/(x+m)+1/(y+m)+1/(z+m)=1/m,那么a,b,c是这个方程的3个根方程化为整式为(m+x)(m+y)(m+z)=[(m+x)(m+y)+(m+y)(m+z)+(m+x)(m+z)]*m整理得2m^3+(x+y+z)m^2-xyz=0,注意到这个方程一次项系数为0所以...