a) N=$\frac{x}{\sqrt{x}-1} - \frac{2x-\sqrt{x}}{x-\sqrt{x}}$
=$\frac{x}{\sqrt{x}-1} - \frac{2\sqrt{x}-1}{\sqrt{x}-1}$
=$\frac{x-(2\sqrt{x}-1)}{\sqrt{x}-1}$
=$\frac{(\sqrt{x}-1)^2}{\sqrt{x}-1}$
=\sqrt{x}-1
b) 2020+2$\sqrt{2019}$ =2019+2$\sqrt{2019}$+1=$(\sqrt{2019}+1)^2$
Vậy N=\sqrt{x}-1=$\sqrt{(\sqrt{2019}+1)^2}$ -1=(\sqrt{2019}+1)-1=\sqrt{2019}