若实数a,b ,c满足abc不等于0,1/a+1/b+1/c=1/(a=b+c),求证a,b,c,至少有两个互为相反数?
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![若实数a,b ,c满足abc不等于0,1/a+1/b+1/c=1/(a=b+c),求证a,b,c,至少有两个互为相反数?](/uploads/image/z/6892870-22-0.jpg?t=%E8%8B%A5%E5%AE%9E%E6%95%B0a%2Cb+%2Cc%E6%BB%A1%E8%B6%B3abc%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C1%2Fa%2B1%2Fb%2B1%2Fc%3D1%2F%28a%3Db%2Bc%29%2C%E6%B1%82%E8%AF%81a%2Cb%2Cc%2C%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%A4%E4%B8%AA%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%3F)
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若实数a,b ,c满足abc不等于0,1/a+1/b+1/c=1/(a=b+c),求证a,b,c,至少有两个互为相反数?
若实数a,b ,c满足abc不等于0,1/a+1/b+1/c=1/(a=b+c),求证a,b,c,至少有两个互为相反数?
若实数a,b ,c满足abc不等于0,1/a+1/b+1/c=1/(a=b+c),求证a,b,c,至少有两个互为相反数?
1/a+1/b+1/c=1/(a+b+c)(bc+ac+ab)/(abc)=1/(a+b+c)(bc+ac+ab)(a+b+c)=abcabc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc=abcb^2c+bc^2+a^2c+ac^2+a^2b+ab^2+2abc=0(b^2c+a^2b+ab^2+abc)+(bc^2+a^2c+ac^2+abc)=0b(bc+a^2+ab+ac)+c(bc+a^2+ac+ab)=0(b+c)(bc+a^2+ab+ac)=0(b+c)[(bc+ab)+(a^2+ac)]=0(b+c)[b(a+c)+a(a+c)]=0(b+c)(b+a)(a+c)=0所以a+b=0或b+c=0或c+a=0,至少有两个为相反数.