\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\\ \Leftrightarrow \frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\\ \Leftrightarrow \frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\\ \Leftrightarrow \frac{x}{15}=\frac{y}{20}=\frac{z}{40}\\\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\\ \Rightarrow x=15k;y=20k;z=40k\\ xy=1200\\ \Leftrightarrow 15k.20k=300k^2=1200\\ \Leftrightarrow k^2=4\)
\(\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ k=2\Rightarrow\left\{{}\begin{matrix}x=15k=15.2=30\\y=20k=20.2=40\\z=40k=40.2=80\end{matrix}\right.\\ k=-2\Rightarrow\left\{{}\begin{matrix}x=15k=15.\left(-2\right)=-30\\y=20k=20.\left(-2\right)=-40\\z=40k=40.\left(-2\right)=-80\end{matrix}\right.\)
\(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)
\(\Rightarrow\dfrac{x-9}{15}=\dfrac{y-12}{20}=\dfrac{z-24}{40}=k\)
\(\Rightarrow\left(15k+9\right)\left(20k+12\right)=1200\)
\(\Rightarrow3.4\left(5k+3\right)\left(5k+3\right)=1200\)
\(\Rightarrow\left(5k+3\right)\left(5k+3\right)=1200:3:4\)
\(\Rightarrow\left(5k+3\right)^2=100\)
\(\Rightarrow\left[{}\begin{matrix}5k+3=10\\5k+3=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}k=\dfrac{7}{5}\\k=\dfrac{-13}{5}\end{matrix}\right.\)
+) Với \(k=\dfrac{7}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}.15+9\\y=\dfrac{7}{5}.20+12\\z=\dfrac{7}{5}.40+24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=80\end{matrix}\right.\)
+) Với \(k=\dfrac{-13}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-13}{5}.15+9\\y=\dfrac{-13}{5}.20+12\\z=\dfrac{-13}{5}.40+24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-30\\y=-40\\z=-80\end{matrix}\right.\)
Vậy .................
