Ta có:
\(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)
\(\Rightarrow\dfrac{x-9}{15}=\dfrac{y-12}{20}=\dfrac{z-24}{40}\)
\(\Rightarrow\dfrac{x}{15}-\dfrac{9}{15}=\dfrac{y}{20}-\dfrac{12}{20}=\dfrac{z}{40}-\dfrac{24}{40}\)
\(\Rightarrow\dfrac{x}{15}-\dfrac{3}{5}=\dfrac{y}{20}-\dfrac{3}{5}=\dfrac{z}{40}-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}\)
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=15k\\y=20k\end{matrix}\right.\)
và \(xy=1200\)
\(\Rightarrow15k.20k=1200\)
\(\Rightarrow300.k^2=1200\)
\(\Rightarrow k^2=4=\left(2\right)^2=\left(-2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+) TH1: \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=40.2=80\end{matrix}\right.\)
+) TH2: \(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.\left(-2\right)=-30\\y=20.\left(-2\right)=-40\\z=40.\left(-2\right)=-80\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\in\left\{\left(30;40;80\right);\left(-30;-40;-80\right)\right\}\)