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