已知a+b+c=0,证明:a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+3=0

这道题 我感觉简单啊，楼主应该不管题怎样，先进行简化整理，再与已知条件联系。方可解出。

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+3=0
进行整理过程为：
(a/b + a/c )+ (b/a + b/c)+(c/a + c/b)+ 3 =0

a/b + a/c + b/a + b/c + c/a + c/b + 3 =0

(a/b + c/b )+ (b/c + a/c)+(c/a + b/a)+ 3 =0

[(a+c)/b ] + [(a+b)/c ]+[(b+c)/a ]+ 3 =0

引用a+b+c=0 也就是a+b=-c,b+c=-a,或a+c=-b代入上式
可得：
[(-b)/b ] + [(-c)/c ]+[(-a)/a ]+ 3 =0

(-1) + (-1) + (-1) +3=0
0=0

即可证明了

2,4

左边：a/b+a/c+b/a+b/c+c/a+c/b=-3
(b+c)/a+(a+c)/b+(a+b)/c=-3
-a/a-b/b-c/c=-3
-1-1-1=-3
即左边等于右边成立

证明：a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+3
=a/b+a/c+b/a+b/c+c/a+c/b+3
=(b+c)/a+(a+c)/b+(a+b)/c+3
又因为a+b+c=0,所以a+b=-c
b+c=-a
a+c=-b
即，-a/a+ -b/b+ -c/c+3=0
所以：a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+3=0

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+3=0
a/b+a/c+b/a+b/c+c/a+c/b+3=0
(a+b)/