Để bth $\sqrt{\frac{1}{2019}}-\frac{2019}{(x-2020)^{2}}$ có nghĩa
⇔$\left \{ {{\frac{1}{2019-x}\geq0} \atop {x-2020\neq0}} \right.$
⇔$\left \{ {{2019-x>0} \atop {x\neq2020}} \right.$
⇔$\left \{ {{-x>-2019} \atop {x\neq2020}} \right.$
⇔$\left \{ {{x<2019} \atop {x\neq2020}} \right.$
Vậy $x<2019,x\neq2020$ thì bth $\sqrt{\frac{1}{2019}}-\frac{2019}{(x-2020)^{2}}$ có nghĩa
⇒Chọn B