已知a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的值
:∵a+b+c=0
∴a+b=-c,或a+c=-b,或b+c=-a
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
=-b/b+(-c/c)+(-a/a)
=-1-1-1
=-3.
求公式[a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3]/a+b+c
求公式[a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3]/a+b+c
数学人气:636 ℃时间:2020-05-22 23:09:22
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