\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{3}\\ \Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{3}{4}}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{3}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{3}{4}}=\dfrac{49}{\dfrac{43}{12}}=\dfrac{588}{43}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{882}{43}\\y=\dfrac{784}{43}\\z=\dfrac{441}{43}\end{matrix}\right.\)
Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{3}\)
⇔ \(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{9} \)
Áp dụng t/c dtsbn, ta có:
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{9}=\dfrac{12(x+y+z)}{18+16+9}=\dfrac{12.49}{43}=\dfrac{588}{43}\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{882}{43}\\y=\dfrac{784}{43}\\z=\dfrac{441}{43}\end{matrix}\right.\)