9b
Hiển nhiên thấy $x \not= 0,y \not= 0,z \not= 0$.
Do đó, $gt⇒x(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=0$
$⇔1+\dfrac{x}{y}+\dfrac{x}{z}=0$
Tương tự, $1+\dfrac{y}{x}+\dfrac{y}{z}=0,1+\dfrac{z}{x}+ \dfrac{z}{y}=0$
Suy ra $\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+ \dfrac{z}{y}=-3(*)$
Lại có: $gt⇒\dfrac{xy+yz+zx}{xyz}=0⇒xy+yz+zx=0$
$⇒(xy+yz+zx)(\dfrac{1}{x^2}+\dfrac{1}{y^2}+ \dfrac{1}{z^2})=0$( do $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2} \not= 0$)
$⇔\dfrac{xy}{z^2}+\dfrac{xz}{y^2}+\dfrac{yz}{x^2}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+ \dfrac{z}{y}=0$
$⇔\dfrac{xy}{z^2}+\dfrac{xz}{y^2}+\dfrac{yz}{x^2}=3$ ( do $(*)$)