Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\). Khi đó ta có:
\(x=2k;2y=3k\Rightarrow y=\dfrac{3k}{2};3z=4k\Rightarrow z=\dfrac{4k}{3}\)
\(\Rightarrow xyz=108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=108\)
\(\Rightarrow\dfrac{24k^3}{6}=108\Rightarrow k^3=27\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=\dfrac{3\cdot3}{2}=\dfrac{9}{2}\\z=\dfrac{4\cdot3}{3}=4\end{matrix}\right.\)
Vậy....
Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\2y=3k\Rightarrow y=\dfrac{3k}{2}\\3z=4k\Rightarrow z=\dfrac{4k}{3}\end{matrix}\right.\)
Mà \(xyz=108\)
\(\Rightarrow2k.\dfrac{3k}{2}.\dfrac{4k}{3}=108\)
\(\Rightarrow2k.\dfrac{3}{2}k.\dfrac{4}{3}k=108\)
\(\Rightarrow k^3.4=108\)
\(\Rightarrow k^3=\dfrac{108}{4}=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=\dfrac{3.3}{2}=4,5\\z=\dfrac{4.3}{3}=4\end{matrix}\right.\)
Vậy \(x=6;y=4,5;z=4\)
