a) Ta có:
$\left ( 6x + 1 \right )^{2020} \geq 0$ với mọi $x$
$\left ( 3y + 1 \right )^{2020} \geq 0$ với mọi $y$
$\left | x + y + 2z^{2} \right | \geq 0$ với mọi $x, y, z$
$\Rightarrow \left ( 6x + 1 \right )^{2020} + \left ( 3y + 1 \right )^{2020} + \left | x + y + 2z^{2} \right | \geq 0$
$\Leftrightarrow \left\{\begin{matrix}
6x + 1 = 0\\
3y + 1 = 0\\
x + y + 2z^{2} = 0
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
x = -\dfrac{1}{6}\\
y = -\dfrac{1}{3}\\
z = ±\dfrac{1}{2}
\end{matrix}\right.$