Đáp án: $P=-1$
Giải thích các bước giải:
Ta có:
$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=0\rightarrow xy+yz+zx=0$
$\rightarrow \dfrac{xy+yz+zx}{xyz}=0\rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
Lại có:
$A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}$
$\rightarrow A=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{1}{x}+\dfrac{1}{y})^3-3\dfrac{1}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{-1}{z})^3-3\dfrac{1}{xy}(\dfrac{-1}{z})+\dfrac{1}{z^3})$
$\rightarrow A=xyz(3\dfrac{1}{xyz})$
$\rightarrow A=3$
$\rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$
$\rightarrow (\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}-4)^{2019}=(-1)^{2019}=-1$
