已知1⼀a+1⼀b+1⼀c=1⼀(a+b+c),求证1⼀a^2019+1⼀b^2019+1⼀c^2019=1⼀(a^2019+b^2019+c^2019)

2025-05-19 23:57:22
推荐回答(1个)
回答1:

1/a+1/b+1/c=1/(a+b+c)
<==>1/a+1/b=1/(a+b+c)-1/c,
<==>(a+b)/(ab)=-(a+b)/[c(a+b+c)],
<==>a=-b,或ab+ac+bc+c^2=0,
<==>a=-b,或b=-c,或c=-a.
当a=-b时a^2019=(-b)^2019=-b^2019,
∴1/a^2019+1/b^2019+1/c^2019=1/(a^2019+b^2019+c^2019),
同理,b=-c,或c=-a上式也成立。