Ta có : \(4x=2y=3z\)
\(\Rightarrow\frac{4x}{12}=\frac{2y}{12}=\frac{3z}{12}\) \(\Leftrightarrow\frac{x}{3}=\frac{y}{6}=\frac{z}{4}\)
Đặt \(\frac{x}{3}=\frac{y}{6}=\frac{z}{4}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=3k\\y=6k\\z=4k\end{cases}}\)
Mà \(2x-3y+z=16\)
\(\Rightarrow2.3k-3.6k+4k=16\)
\(\Leftrightarrow6k-18k+4k=16\)
\(\Leftrightarrow k.\left(6-18+4\right)=16\)
\(\Leftrightarrow-8k=16\)
\(\Leftrightarrow k=-2\)
\(\Rightarrow\hept{\begin{cases}x=3k=-6\\y=6k=-12\\z=4k=-8\end{cases}}\)
Vậy ...
\(4x=2y=3z\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
Theo tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x}{4}=\frac{3y}{9}=\frac{2x-3y+z}{4-9+4}=\frac{16}{-1}=-16\)
x/2=-16 => x=-16.2=-32
y/3=-16 => y=-16.3=-48
z/4=-16 => z=-16.4=-64
Vậy x=-32 ; y=-48 ; z=-64
