1956波兰竞赛题1/a+1/b+1/c=1/(a+b+c),求证:1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 06:00:45
![1956波兰竞赛题1/a+1/b+1/c=1/(a+b+c),求证:1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)](/uploads/image/z/14436848-56-8.jpg?t=1956%E6%B3%A2%E5%85%B0%E7%AB%9E%E8%B5%9B%E9%A2%981%2Fa%2B1%2Fb%2B1%2Fc%3D1%2F%28a%2Bb%2Bc%29%2C%E6%B1%82%E8%AF%81%EF%BC%9A1%2Fa%5E%282n%2B1%29%2B1%2Fb%5E%282n%2B1%29%2B1%2Fc%5E%282n%2B1%29%3D1%2F%5Ba%5E%282n%2B1%29%2Bb%5E%282n%2B1%29%2Bc%5E%282n%2B1%29%5D%3D1%2F%28a%2Bb%2Bc%29%5E%282n%2B1%29%3D%281%2Fa%2B1%2Fb%2B1%2Fc%29%5E%282n%2B1%29)
1956波兰竞赛题1/a+1/b+1/c=1/(a+b+c),求证:1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)
1956波兰竞赛题
1/a+1/b+1/c=1/(a+b+c),求证:1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)
1956波兰竞赛题1/a+1/b+1/c=1/(a+b+c),求证:1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)
证明:因为1/a+1/b+1/c=1/(a+b+c),所以1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1),这个好证明!
(1/a+1/b+1/c)x(a+b+c),=1化简可得:b/a+a/b+c/a+b/c+c/b+a/c=-2,如果a,b,c为同号.那么上是不满足.用均值不等,可得不能成立,所以a,b,c中有正有负,
假设a>0(随便假设都一样)b<0.那么c<0现在设1/b+1/c=x b+c=y这样题中条件可化简为1/a+x=1/(a+y),这不可能,b>0.那么c<0,那么当x=y=0能够满足,并且不管怎么变化都满足,如果如果x不等于0.那么又和上面的一样,不能成立,
所以a,b,c中一定有两个加起来为0.假设b+c=0那么1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/a^(2n+1),1/(a+b+c)^(2n+1)=
1/a^(2n+1),
1/b+1/c,所以1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/a^(2n+1),
所以1/a^(2n+1)+1/b^(2n+1)+1/c^(2n+1)=1/[a^(2n+1)+b^(2n+1)+c^(2n+1)]=1/(a+b+c)^(2n+1)=(1/a+1/b+1/c)^(2n+1)