已知a+b+c=0,且a,b,c都不等于0,试说明:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
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![已知a+b+c=0,且a,b,c都不等于0,试说明:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0](/uploads/image/z/2521598-14-8.jpg?t=%E5%B7%B2%E7%9F%A5a%2Bb%2Bc%3D0%2C%E4%B8%94a%2Cb%2Cc%E9%83%BD%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E8%AF%95%E8%AF%B4%E6%98%8E%EF%BC%9Aa%281%2Fb%2B1%2Fc%29%2Bb%281%2Fc%2B1%2Fa%29%2Bc%281%2Fa%2B1%2Fb%29%2B3%3D0)
已知a+b+c=0,且a,b,c都不等于0,试说明:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
已知a+b+c=0,且a,b,c都不等于0,试说明:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
已知a+b+c=0,且a,b,c都不等于0,试说明:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
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=-3,
[a+b+c=0,即a+c=-b,余下同理.].
在 a+b+c=0 的两边同除以 a 得 1+b/a+c/a=0 ,
所以 b/a+c/a= -1 ,
同理 a/b+c/b=a/c+b/c= -1 ,
因此,原式左边=a/b+a/c+b/c+b/a+c/a+c/b+3
=(a/b+c/b)+(a/c+b/c)+(b/a+c/a)+3
= -1-1-1+3=0 。