Giải thích các bước giải:
Ta có:
$\dfrac1a+\dfrac1b+\dfrac1c=\dfrac{1}{a+b+c}$
$\to \dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}$
$\to (a+b+c)(ab+bc+ca)=abc$
$\to ((a+b)+c)(ab+c(a+b))=abc$
$\to (a+b)\cdot ab+c(a+b)^2+abc+c^2(a+b)=abc$
$\to (a+b)\cdot ab+c(a+b)^2+c^2(a+b)=0$
$\to (a+b)\cdot (ab+c(a+b)+c^2)=0$
$\to (a+b)\cdot (c+a)\cdot (c+b)=0$
$\to a+b=0$ hoặc $a+c=0$ hoặc $b+c=0$
Không mất tính tổng quát giả sử $b+c=0\to b=-c$
$\to \dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{c^{2017}}=\dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{(-b)^{2017}}$
$\to \dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{c^{2017}}=\dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}-\dfrac{1}{b^{2017}}$
$\to \dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{c^{2017}}=\dfrac{1}{a^{2017}}+0$
$\to \dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{c^{2017}}=\dfrac{1}{a^{2017}}$
Lại có:
$\dfrac{1}{a^{2017}+b^{2017}+c^{2017}}=\dfrac{1}{a^{2017}+b^{2017}+(-b)^{2017}}$
$\to \dfrac{1}{a^{2017}+b^{2017}+c^{2017}}=\dfrac{1}{a^{2017}+b^{2017}-b^{2017}}$
$\to \dfrac{1}{a^{2017}+b^{2017}+c^{2017}}=\dfrac{1}{a^{2017}+0}$
$\to \dfrac{1}{a^{2017}+b^{2017}+c^{2017}}=\dfrac{1}{a^{2017}}$
$\to \dfrac{1}{a^{2017}}+\dfrac{1}{b^{2017}}+\dfrac{1}{c^{2017}}= \dfrac{1}{a^{2017}+b^{2017}+c^{2017}}$
