Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y)(z+x)(z+y)=0$
Nếu $x+y=0\Rightarrow x=-y$
Khi đó:
$\frac{1}{x^{2023}}+\frac{1}{y^{2023}}+\frac{1}{z^{2023}}=\frac{(-y)^{2023}}+\frac{1}{y^{2023}}+\frac{1}{z^{2023}}=\frac{1}{z^{2023}}$
$=\frac{1}{(-y)^{2023}+y^{2023}+z^{2023}}=\frac{1}{x^{2023}+y^{2023}+z^{2023}}$ (đpcm)
Tương tự với $y+z=0; z+x=0$
Ta có đpcm.
