\(3x=4y=2z\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}\)
Áp dụng t/d dtsbn:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}=\dfrac{2x}{8}=\dfrac{2x+y-z}{8+3-6}=\dfrac{-5}{5}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).3=-3\\z=\left(-1\right).6=-6\end{matrix}\right.\)
\(3x=4y=2z\)
⇒\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}\)
⇒\(\dfrac{2x}{8}=\dfrac{y}{3}=\dfrac{z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{8}=\dfrac{y}{3}=\dfrac{z}{6}=\dfrac{2x+y-z}{8+3-6}=\dfrac{-5}{5}=-1\)
⇒\(\left\{{}\begin{matrix}x=-1.4=-4\\y=-1.3=-3\\z=-1.6=-6\end{matrix}\right.\)