\(2x=3y=4z\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{3x}{18}=\dfrac{2y}{8}=\dfrac{z}{3}=\dfrac{3x-2y+z}{18-8+3}=\dfrac{26}{13}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.2=12\\y=4.2=8\\z=3.2=6\end{matrix}\right.\)
2x=3y=4z\(\Leftrightarrow\left\{{}\begin{matrix}x=2z\\y=\dfrac{4}{3}z\end{matrix}\right.\)
ta có \(3x-2y+z=26\)
\(3.2z-2.\dfrac{4}{3}z+z=26\)
\(\dfrac{13}{3}z=26\)
\(z=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.6=12\\y=\dfrac{4}{3}.6=8\end{matrix}\right.\)
ta có:
\(\text{2x=3y=4z}\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}\\ =\dfrac{3x}{12}=\dfrac{2y}{6}=\dfrac{z}{2}=\dfrac{3x-2y+3z}{12-6+2}=\dfrac{26}{8}=\dfrac{13}{4}\\ \Rightarrow\left\{{}\begin{matrix}x=13\\y=\dfrac{39}{4}\\z=\dfrac{13}{2}\end{matrix}\right.\)
