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`

Đáp án:

\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3 \to \bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)^2=9\]

\[\dfrac{2}{xy}-9=\dfrac{1}{z^2}\]

\[\to \dfrac{2}{xy}- \bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)^2=\dfrac{1}{z^2}\] \[\leftrightarrow \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}+\dfrac{1}{z^2}=0\] \[\leftrightarrow \bigg(\dfrac{1}{x}+\dfrac{1}{z}\bigg)^2+\bigg(\dfrac 1z+\dfrac 1y\bigg)^2=0\]

mà $\bigg(\dfrac{1}{x}+\dfrac{1}{z}\bigg)^2+\bigg(\dfrac 1z+\dfrac 1y\bigg)^2\ge 0$
$→$ Đẳng thức xảy ra $↔x=y=-z$

$→x=y=\dfrac{1}{3}; z=-\dfrac{1}{3}$

Thay vào $P=(x+3y+z)^{2019}=1$

Vậy $P=1$