Ta có:
`\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3 ⇒ ` `(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2=9`
Do đó:
`\frac{2}{xy}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2=\frac{1}{z^2}`
$⇔$ `(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2=\frac{2}{xy}-\frac{1}{z^2}`
$⇔$ `\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}-\frac{2}{xy}+\frac{1}{z^2}=0`
$⇔$ `(\frac{1}{x}+\frac{1}{z})^2+(\frac{1}{y}+\frac{1}{z})^2=0`
$⇔$ $\left \{ {{(\frac{1}{x}+\frac{1}{z})^2=0} \atop {(\frac{1}{y}+\frac{1}{z})^2=0}} \right.$
$⇔$ $\left \{ {{\frac{1}{x}=\frac{1}{-z}} \atop {\frac{1}{y}=\frac{1}{-z}}} \right.$
$⇔$ `x=y=-z`
Thay vào `\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3` ta được: `x=y=\frac{1}{3};z=-\frac{1}{3}`
Khi đó: `P=(\frac{1}{3}+3.\frac{1}{3}+\frac{-1}{3})^2019=1^{2019}=1`