Đáp án: $P=-1$
Giải thích các bước giải:
Ta có:
$x^2+y^2+z^2=9$
$\to x^2+y^2+z^2=\left(x+y+z\right)^2$ vì $x+y+z=3$
$\to x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)$
$\to xy+yz+zx=0$
$\to \dfrac{xy+yz+zx}{xyz}=0$
$\to \dfrac1x+\dfrac1y+\dfrac1z=0$
$\to \dfrac1z=-\left(\dfrac1x+\dfrac1y\right)$
$\to \left(\dfrac1z\right)^3=-\left(\dfrac1x+\dfrac1y\right)^3$
$\to \dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3.\dfrac{1}{x^2}.\dfrac{1}{y}+3.\dfrac{1}{x}.\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)$
$\to\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
$\to \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{xyz}$
$\to xyz\left(\dfrac1{x^3}+\dfrac1{y^3}+\dfrac1{z^3}\right)=3$
$\to \dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=3$
$\to P=\left(3-4\right)^{2019}$
$\to P=-1$
