Ta có: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2019}\)
⇔ \(\frac{x+y}{xy}=\frac{1}{2019}\)
\(\Rightarrow\left\{{}\begin{matrix}x-2019=\frac{2019x}{y}\\y-2019=\frac{2019y}{x}\end{matrix}\right.\)
Xét: \(\sqrt{x-2019}+\sqrt{y-2019}\)
= \(\sqrt{\frac{2019x}{y}}+\sqrt{\frac{2019y}{x}}\)
= \(\sqrt{2019}\left(\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}}\right)\)
= \(\sqrt{2019}.\frac{x+y}{xy}\)
P = \(\frac{\sqrt{x+y}}{\sqrt{x-2019}+\sqrt{y-2019}}\)
= \(\frac{\sqrt{x+y}}{\frac{\sqrt{2019}.\left(x+y\right)}{\sqrt{xy}}}\)
= \(\sqrt{x+y}.\frac{\sqrt{xy}}{\sqrt{2019}.\left(x+y\right)}\)
= \(\frac{\sqrt{xy}}{\sqrt{2019}.\sqrt{x+y}}\)
= \(\frac{\sqrt{2019}}{\sqrt{2019}}\)
= 1
Vậy P = 1
