\(\dfrac{x}{0,3}=\dfrac{y}{0,2}=2z=\dfrac{3x}{0,9}=\dfrac{z}{\dfrac{1}{2}}=\dfrac{z-3x}{\dfrac{1}{2}-0,9}=\dfrac{1}{-\dfrac{2}{5}}=-\dfrac{5}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5.0,3}{2}=-\dfrac{3}{4}\\y=-\dfrac{5.0,2}{2}=-\dfrac{1}{2}\\z=-\dfrac{5}{2.2}=-\dfrac{5}{4}\end{matrix}\right.\)
Ta có: \(\dfrac{x}{0.3}=\dfrac{y}{0.2}=\dfrac{2z}{1}\)
nên \(\dfrac{3x}{0.9}=\dfrac{y}{0.2}=\dfrac{z}{0.5}\)
mà z-3x=1
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
Δ\(\dfrac{3x}{0.9}=\dfrac{y}{0.2}=\dfrac{z}{0.5}=\dfrac{z-3x}{0.5-0.9}=\dfrac{1}{0.4}=\dfrac{5}{2}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\cdot\dfrac{3}{10}=\dfrac{3}{4}\\y=\dfrac{5}{2}\cdot\dfrac{1}{5}=\dfrac{1}{2}\\z=\dfrac{5}{2}:2=\dfrac{5}{4}\end{matrix}\right.\)
