已知abc不等于0,a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的值

已知abc不等于0,a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的值
已知abc不等于0,a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的值

已知abc不等于0,a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的值
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=a/b+a/c+b/c+b/a+c/a+c/b+1+1+1
=[(a+b)/c+c/c]+[(a+c)/b+b/b]+[(b+c)/a+a/a]
=(a+b+c)/c+(a+b+c)/b+(a+b+c)/a
=0+0+0
=0

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

由a+b+c=0，-C=A+B，-B=A+C，-A=B+C
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=A/B+A/C+B/C+B/A+C/A+C/B+3
=（B+C）/A+（A+B）/B+（A+B）/C+3
=-1-1-1+3
=0

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

化简等式有a/b+a/c+b/c+b/a+c/a+c/b+3,和并等式有(a+c)/b+(a+b)/c+(b+c)/a+3，因为a+b+c=0,所以a+c=-b,a+b=-c,b+c=-a,所以等式等于-b/b-c/c-a/a+3=-1-1-1+3=0，所以答案为0

因为a+b+c=0，所以a+b=﹣c，a+c=﹣b，b+c=﹣a，，又abc≠0
所以a（1/b+1/c）+b（1/c+1/a）+c（1/a+1/b）=a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)/a=(﹣b/b)+(﹣c/c)+(﹣a/a)=﹣3