Giải thích các bước giải:
Ta có:
$x+y+z=0$
$\to \dfrac{x+y+z}{xyz}=0$
$\to \dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy}=0$
$\to 2(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy})=0$
$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy})=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$
$\to (\dfrac1x+\dfrac1y+\dfrac1z)^2=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$
$\to \sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{(\dfrac1x+\dfrac1y+\dfrac1z)^2}$
$\to \sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=|\dfrac1x+\dfrac1y+\dfrac1z|$
