22. Given $x \leq 0$, if $a-x=2019, b+|x|=2020, c+\sqrt{x^{2}}=2021$. If $a b c=24$, then $\sqrt[3]{\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}}$ is
Answer: $\frac{1}{2}$