Question

CM : neu   \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)  thi \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}\)

Answer 1

ĐK: a,b,c khác 0 và a+b+c khác 0

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\Rightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\Rightarrow\left(a+b\right)\left(\frac{c^2+ca+ab+bc}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(a+b+c\right)}=0\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> hoặc a=-b hoặc b=-c hoặc c=-a.

Khi đó đẳng thức:

\(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}\)đúng với mọi lũy thừa lẻ 2n+1. ĐPCM.

Answer 2

Khó zữ

Answer 3

Khó zữ zậy nó lang ngoang làm sao ý.

Answer 4

bó tay