\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{3y}{6}=\dfrac{3z}{9}=\dfrac{x-3y+3z}{5+6-9}=\dfrac{24}{2}=12\\ \Rightarrow\left\{{}\begin{matrix}x=60\\y=24\\z=36\end{matrix}\right.\)
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{3y}{6}=\dfrac{3z}{9}\)
Áp dụng t/c DTSBN, ta có:
\(\dfrac{x}{5}=\dfrac{3y}{6}=\dfrac{3z}{9}=\dfrac{x-3y+3z}{5-6+9}=\dfrac{24}{8}=3\)
\(\dfrac{x}{5}=3\Rightarrow x=15\)
\(\dfrac{y}{2}=3\Rightarrow y=6\)
\(\dfrac{z}{3}=3\Rightarrow=9\)
Vậy...
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}\text{ và }x-3y+3z=24\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{x-3y+3z}{5-3.2+3.3}=\dfrac{24}{8}=3\)
\(\Rightarrow x=3.5=15\)
\(y=3.2=6\)
\(z=3.3=9\)
