Question

已知a+b+c=0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)的值

Answer 1

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

Answer 2

解：
a（1/b+1/c)+b(1/a+1/c)+c(1/b+1/a）
=a/b+a/c+a/a+b/a+b/c+b/b+c/b+c/a+c/c -3
=a/b+b/b+c/b+a/a+a/b+a/c+a/c+b/c+c/c -3
=(a+b+c)/b+(a+b+c)/a+(a+b+c)/c -3
=0/b+0/a+0/c -3
=-3

引用的

Answer 3

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

Answer 4

原式=(b+c)/a+(c+a)/b+(a+b)/c=-a/a+(-b/b)+(-c/c)=-3