ĐKXĐ : \(x;y;z\ne0\)
Vì \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) nên ta có :
\(A=\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{x+z}{y}=x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(=x.\dfrac{-1}{x}+y.\dfrac{-1}{y}+z.\dfrac{-1}{z}=-1.3=-3\)