已知a^2+b^2+c^2=(a+b+c)^2且abc≠0,求1/a+1/b+1/c的值

(a+b+c)^2
=(a+b)^2+2c(a+b)+c^2
=a^2+2ab+b^2+2ac+2bc+c^2
＝a^2+b^2+c^2+2ab+2bc+2ac
=a^2+b^2+c^2
所以2ab+2bc+2ac=0
ab+bc+ac=0
所以1/a+1/b+1/c
=(bc+ac+ab/)abc
=0