\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\) và \(x+y+z=49\) Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=12.\dfrac{3}{2}=18\) \(\Rightarrow\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=12.\dfrac{4}{3}=16\) \(\Rightarrow\dfrac{z}{\dfrac{5}{4}}=12\Rightarrow z=12.\dfrac{5}{4}=15\) Vậy \(\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)