分式加减 (22 17:21:41)abc不等于0,a+b+c=0,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)等于多少?
问题描述：
分式加减 (22 17:21:41)
abc不等于0,a+b+c=0,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)等于多少?
答
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a(b+c)/(bc)+b(a+c)/(ac)+c(a+b)/(ab)
=-a^3/(abc)-b^3/(abc)-c^3/(abc)
=((b+c)^3-b^3-c^3)/(-bc(b+c))
=(3b^2c+3bc^2)/(-bc(b+c))
=-3
答
=a/b+a/c+b/a+b/c+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)/a
=-3
答
-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因为 abc不等于0,a+b+c=0所以 a+c=-ba+b=-cb+c=-a所以(a+c)/b+(a+b)/c+(b+c)/a=-1+（-1）+(-1)=-3即a(1/b+1/c)+b(1/a+1/c...