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

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

已知a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值
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/b+c/b)+(a/c+b/c)+(b/a+c/a)
=(a+c)/b+(a+b)/c+(b+c)/a(将a+c=-b,a+b=-c,b+c=-a代人)
=-b/b--c/c-a/a
=-1-1-1
=-3

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

FJNBNMJ

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=abc*[a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)]/abc............(1)
=[a^2*(b+c)+b^2*(a+c)+c^2*(a+b)]/abc...............(...

全部展开

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=abc*[a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)]/abc............(1)
=[a^2*(b+c)+b^2*(a+c)+c^2*(a+b)]/abc...............(2)
=[ab(a+b)+ac(a+c)+bc(b+c)]/abc.............(3)
=-3abc/abc=-3
注：第一步是讲分子分母同乘以abc,第二步后将各项重新组合就可以得到第三步；然后利用a=-b-c
,b=-a-c,c=-a-b,就能得到答案

收起