Đáp án:
Giải thích các bước giải:
Có: $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$⇔ \dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}$
$⇔ (a+b+c)(ab+bc+ca)=abc$
$⇔ a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+2abc=0$
$⇔ ab(a+b)+c^2(a+b)+c(a^2+2ab+b^2)=0$
$⇔ ab(a+b)+c^2(a+b)+c(a+b)^2=0$
$⇔ (a+b)(ab+ac+bc+c^2)=0$
$⇔ (a+b)(b+c)(c+a)=0$
$⇔ \left[ \begin{array}{l}a=-b\\b=-c\\c=-a\end{array} \right.$
Nếu $a=-b ⇔ a+b=0$ thì:
$\dfrac{1}{a^{2021}}+\dfrac{1}{b^{2021}}+\dfrac{1}{c^{2021}}$
$=\dfrac{1}{-b^{2021}}+\dfrac{1}{b^{2021}}+\dfrac{1}{(a+b+c)^{2021}}$
$=\dfrac{1}{(a+b+c)^{2021}}$
Tương tự với $b=-c; c=-a$.
Vậy ta có $đpcm$
