$\dfrac1x +\dfrac1y +\dfrac1z =\dfrac{1}{x+y+z}$
$\to \left(\dfrac1x +\dfrac1y\right) +\left(\dfrac1z -\dfrac{1}{x+y+z}\right)= 0$
$\to \dfrac{x+y}{xy} +\dfrac{x+y}{z(x + y + z)}=0$
$\to (x+y)\cdot\dfrac{z(x + y + z) + xy}{xyz(x + y + z)} = 0$
$\to (x+y)\cdot\dfrac{zx + z^2 + zy + xy}{xyz(x + y + z)}=0$
$\to \dfrac{(x+y)(y + z)(z + x)}{xyz(x+y+z)}=0$
$\to \left[\begin{array}{l}x + y = 0\\y + z = 0 \\z + x = 0\end{array}\right.$
$\to \left[\begin{array}{l}x = -y\\y = - z\\z = - x\end{array}\right.$
$+)\quad x = -y$
$\to \begin{cases}\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} =\dfrac{1}{z^{2019}}\\\dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{z^{2019}}\end{cases}$
$\to \dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{z^{2019}}$
$+)\quad y = - z$
$\to \begin{cases}\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} =\dfrac{1}{x^{2019}}\\\dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{x^{2019}}\end{cases}$
$\to \dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{x^{2019}}$
$+)\quad z = - x$
$\to \begin{cases}\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} =\dfrac{1}{y^{2019}}\\\dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{y^{2019}}\end{cases}$
$\to \dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019} +y^{2019} + z^{2019}}=\dfrac{1}{y^{2019}}$
Vậy $\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019} +y^{2019} + z^{2019}}$
