已知a+b+c=0,且a、b、c都不等于零,求证: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都不等于零,求证:a(1/b+1/c)+b(1/c+1/a)=c(1/a+1/b)+3=0](/uploads/image/z/8706708-36-8.jpg?t=%E5%B7%B2%E7%9F%A5a%2Bb%2Bc%3D0%2C%E4%B8%94a%E3%80%81b%E3%80%81c%E9%83%BD%E4%B8%8D%E7%AD%89%E4%BA%8E%E9%9B%B6%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%EF%BC%881%2Fb%2B1%2Fc%29%2Bb%281%2Fc%2B1%2Fa%29%3Dc%281%2Fa%2B1%2Fb%29%2B3%3D0)
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已知a+b+c=0,且a、b、c都不等于零,求证: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都不等于零,求证: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都不等于零,求证:a(1/b+1/c)+b(1/c+1/a)=c(1/a+1/b)+3=0
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+3
=(a+b)/c+(a+c)/b+(b+c)/a+3
=-c/c+(-b)/b+(-a)/a+3
=-1-1-1+3
=0