mình nghĩ bạn chép sai đề bài
dấu ''='' thứ 2 thay bằng dấu ''+''
ta có
\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\)
\(\Rightarrow19\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
lại có
\(\dfrac{7x}{y+z}+\dfrac{7y}{x+z}+\dfrac{7z}{x+y}=\dfrac{133}{10}\)
\(\Rightarrow7\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{19}{10}\)
\(\Rightarrow\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}+\dfrac{x+y+z}{x+y}=\dfrac{49}{10}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{49}{10}\)
\(\Rightarrow\dfrac{7}{10}\left(x+y+z\right)=\dfrac{49}{10}\Rightarrow\left(x+y+z\right)^2=49.\)
