Giải thích các bước giải:
Ta có :
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
$\to (\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})^3=0$
$\to \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3(\dfrac{1}{x}+\dfrac{1}{y})(\dfrac{1}{y}+\dfrac{1}{z})(\dfrac{1}{z}+\dfrac{1}{x})=0$
$\to \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3.\dfrac{-1}{z}.\dfrac{-1}{x}.\dfrac{-1}{y}=0$
$\to \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}$
Lại có :
$P=\dfrac{yz}{3x^2}+\dfrac{xz}{3y^2}+\dfrac{xy}{3z^2}$
$\to P=\dfrac{xyz}{3}(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$
$\to P=\dfrac{xyz}{3}.\dfrac{3}{xyz}=1$
