theo đề bài ta có: \(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\) và 4x-3y+2z=32
=>\(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}\)
Áp dụng dãy tỉ số bằng nhau ta đc:
\(\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{32}{4}=8\)
=>4x=8.4=32=>x=8
3y=8.6=48=>y=16
2z=8.6=48=>z=24
vậy x=8
y=16
z=24
\(x=\dfrac{y}{2}=\dfrac{z}{3}\&4x-3y+2z=32\)
Từ \(x=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{4x}{4}=\dfrac{3y}{2.3}=\dfrac{2z}{2.3}\Rightarrow\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{32}{4}=8\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4x}{4}=8\\\dfrac{3y}{6}=8\\\dfrac{2z}{6}=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\\dfrac{y}{2}=8\\\dfrac{z}{3}=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=16\\z=24\end{matrix}\right.\)
