Giải:
\(2x=5y=6z\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{10}=\dfrac{3z}{18}_{\left(1\right)}\)và \(3x-z+2y=24_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{2}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{3z-x+2y}{18-2+10}=\dfrac{24}{26}=\dfrac{12}{13}.\)
Từ đó:
\(\dfrac{x}{2}=\dfrac{12}{13}\Rightarrow x=\dfrac{2.12}{13}=\dfrac{24}{13}.\)
\(\dfrac{2y}{10}=\dfrac{12}{13}\Rightarrow2y=\dfrac{10.12}{13}=\dfrac{120}{13}\Rightarrow y=\dfrac{60}{13}.\)
\(\dfrac{3z}{18}=\dfrac{12}{13}\Rightarrow3x=\dfrac{18.12}{13}=\dfrac{216}{13}\Rightarrow z=\dfrac{72}{13}.\)
Vậy: \(\left[{}\begin{matrix}x=\dfrac{24}{13}.\\y=\dfrac{60}{13}.\\z=\dfrac{72}{13}.\end{matrix}\right.\)
