实数a、b、c满足1/a+1/b+1/c=1/(a+b+c).求证:1/a7+1/b7+1/c7=1/(a7+b7+c7)注:7均为七次方
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 04:31:44
![实数a、b、c满足1/a+1/b+1/c=1/(a+b+c).求证:1/a7+1/b7+1/c7=1/(a7+b7+c7)注:7均为七次方](/uploads/image/z/1069837-61-7.jpg?t=%E5%AE%9E%E6%95%B0a%E3%80%81b%E3%80%81c%E6%BB%A1%E8%B6%B31%2Fa%2B1%2Fb%2B1%2Fc%3D1%2F%28a%2Bb%2Bc%29.%E6%B1%82%E8%AF%81%EF%BC%9A1%2Fa7%2B1%2Fb7%2B1%2Fc7%3D1%2F%28a7%2Bb7%2Bc7%29%E6%B3%A8%EF%BC%9A7%E5%9D%87%E4%B8%BA%E4%B8%83%E6%AC%A1%E6%96%B9)
实数a、b、c满足1/a+1/b+1/c=1/(a+b+c).求证:1/a7+1/b7+1/c7=1/(a7+b7+c7)注:7均为七次方
实数a、b、c满足1/a+1/b+1/c=1/(a+b+c).
求证:1/a7+1/b7+1/c7=1/(a7+b7+c7)
注:7均为七次方
实数a、b、c满足1/a+1/b+1/c=1/(a+b+c).求证:1/a7+1/b7+1/c7=1/(a7+b7+c7)注:7均为七次方
证明:
1/a+1/b+1/c=1/(a+b+c)
(通分)
(bc+ac+ab)/(abc)=1/(a+b+c)
(bc+ac+ab)(a+b+c)=abc
abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc=abc
b^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)=0
b(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,
即a=-b或b=-c或c=-a.
所以1/a7+1/b7+1/c7=1/(a7+b7+c7)