$\\$
Đặt `(2x)/3 = (3y)/4 =(5z)/6 =k (k \ne 0)`
`->` $\begin{cases} \dfrac{2x}{3}=k\\\dfrac{3y}{4}=k\\\dfrac{5z}{6}=k \end{cases}$ `->` $\begin{cases} 2x=3k\\3y=4k\\5z=6k \end{cases}$ `->` $\begin{cases} x=\dfrac{3}{2}k\\y=\dfrac{4}{3}k\\z=\dfrac{6}{5}k\end{cases}$
Có : `x^2 . y^2 . z^2 =144/25`
`-> (3/2k)^2 . (4/3k)^2 . (6/5k)^2=144/25`
`-> 9/4k^2 . 16/9 k^2 . 36/25k^2 = 144/25`
`-> (9/4 . 16/9 . 36/25) (k^2 . k^2 . k^2) = 144/25`
`-> 144/25 . k^6 = 144/25`
`-> k^6 = 144/25 : 144/25`
`-> k^6=1`
`->` \(\left[ \begin{array}{l}k^6=1^6\\k^6=(-1)^6\end{array} \right.\) `->` \(\left[ \begin{array}{l}k=1\\k=-1\end{array} \right.\) (Thỏa mãn)
Với `k=1`
`->` $\begin{cases} x=\dfrac{3}{2}.1\\y=\dfrac{4}{3}.1\\z=\dfrac{6}{5}.1\end{cases}$ `->` $\begin{cases} x=\dfrac{3}{2}\\y=\dfrac{4}{3}\\z=\dfrac{6}{5}\end{cases}$
Với `k=-1`
`->` $\begin{cases} x=\dfrac{3}{2}.(-1)\\y=\dfrac{4}{3}.(-1)\\z=\dfrac{6}{5}.(-1)\end{cases}$ `->` $\begin{cases} x=\dfrac{-3}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-6}{5}\end{cases}$
Vậy `(x;y;z) = (3/2; 4/3; 6/5), ( (-3)/2; (-4)/3; (-6)/5)`
