Đáp án: $A=-1$
Giải thích các bước giải:
Ta có:
$2\left(xy+yz+zx\right)=\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)=3^2-9=0$
$\to xy+yz+zx=0$
$\to\dfrac{xy+yz+zx}{xyz}=0$
$\to\dfrac1x+\dfrac1y+\dfrac1z=0$
$\to \dfrac1z=-\left(\dfrac1x+\dfrac1y\right)$
$\to\left(\dfrac1z\right)^3=-\left(\dfrac1x+\dfrac1y\right)^3$
$\to\left(\dfrac1z\right)^3=-\left(\left(\dfrac1x\right)^3+\left(\dfrac1y\right)^3+3.\dfrac1x.\dfrac1y\left(\dfrac1x+\dfrac1y\right)\right)$
$\to\left(\dfrac1z\right)^3=-\left(\left(\dfrac1x\right)^3+\left(\dfrac1y\right)^3+3.\dfrac1x.\dfrac1y\left(-\dfrac1z\right)\right)$
$\to\dfrac1{z^3}=-\left(\dfrac1{x^3}+\dfrac1{y^3}-3.\dfrac1x.\dfrac1y\dfrac1z\right)$
$\to\dfrac1{z^3}=-\dfrac1{x^3}-\dfrac1{y^3}+3.\dfrac1x.\dfrac1y\dfrac1z$
$\to\dfrac1{x^3}+\dfrac1{y^3}+\dfrac1{z^3}=3.\dfrac1x.\dfrac1y\dfrac1z$
$\to\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=3$
$\to A=(3-4)^{2021}$
$\to A=-1$
