\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\\ \Rightarrow19.\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\\ \Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
\(\dfrac{7x}{y+z}+\dfrac{7z}{x+y}+\dfrac{7y}{x+z}=\dfrac{133}{10}\\ \Rightarrow\dfrac{x}{y+z}+\dfrac{z}{x+y}+\dfrac{y}{x+z}=\dfrac{133}{10}:7=\dfrac{19}{10}\\ \Rightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{z}{x+y}+1\right)+\left(\dfrac{y}{x+z}+1\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right).\dfrac{7}{10}=\dfrac{49}{10}\\ \Rightarrow x+y+z=7\)
