\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{24}{z-24}\Rightarrow\)\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)

Do x . y = 1200 => ( 15k + 9) ( 20k + 12) = 1200

3(5k+3).4(5k+3) = 1200

12 ( 5k+3)² = 1200

( 5k+3)² =  100 hoặc ( 5k+3)² =  -100

=> 5k+3 = 10 hoặc 5k+3 = -10

=> 5k = 7 hoặc 5k = -13

=> k = 7/5 hoặc k = -13/5

 Vậy

\(\hept{\begin{cases}\text{x = 15 . \frac{7}{5}+9 = 30}\\y=20.\frac{7}{5}+12=40\\z=40.\frac{7}{5}+24=80\end{cases}}\)\(\hept{\begin{cases}x=15.\frac{7}{5}+9=30\\y=20.\frac{7}{5}+12=40\\z=40.\frac{7}{5}+24=80\end{cases}}\)hoặc\(\hept{\begin{cases}x=15.\frac{-13}{5}+9=-30\\y=20.\frac{-13}{5}+12=-40\\z=40.\frac{-13}{5}+24=-80\end{cases}}\)

\(\frac{15}{x-9}\)=\(\frac{5}{\frac{x}{3}-3}\) 

\(\frac{20}{y-12}\)= \(\frac{5}{\frac{y}{4}-3}\)

=> \(\frac{x}{3}\) = \(\frac{y}{4}\)

vậy x = 30, y = 40 ,  z = 80

X=30 Y=40 Z=80

\(\frac{15}{x-9}=\frac{20}{y-12}\Leftrightarrow15\left(y-12\right)=20\left(x-9\right)\Leftrightarrow15y-180=20x-180\Leftrightarrow15y=20x\Leftrightarrow\frac{y}{20}=\frac{x}{15}\Leftrightarrow\frac{xy}{20}=\frac{x^2}{15}\Leftrightarrow x^2=\frac{15.1200}{20}=900\Leftrightarrow\orbr{\begin{cases}x=30\\x=-30\end{cases}}\)

Chia từng trường hợp tìm y, z.

\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\\ \Leftrightarrow \frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\\ \Leftrightarrow \frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\\ \Leftrightarrow \frac{x}{15}=\frac{y}{20}=\frac{z}{40}\\\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\\ \Rightarrow x=15k;y=20k;z=40k\\ xy=1200\\ \Leftrightarrow 15k.20k=300k^2=1200\\ \Leftrightarrow k^2=4\)

\(\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ k=2\Rightarrow\left\{{}\begin{matrix}x=15k=15.2=30\\y=20k=20.2=40\\z=40k=40.2=80\end{matrix}\right.\\ k=-2\Rightarrow\left\{{}\begin{matrix}x=15k=15.\left(-2\right)=-30\\y=20k=20.\left(-2\right)=-40\\z=40k=40.\left(-2\right)=-80\end{matrix}\right.\)

\(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)

\(\Rightarrow\dfrac{x-9}{15}=\dfrac{y-12}{20}=\dfrac{z-24}{40}=k\)

\(\Rightarrow\left(15k+9\right)\left(20k+12\right)=1200\)

\(\Rightarrow3.4\left(5k+3\right)\left(5k+3\right)=1200\)

\(\Rightarrow\left(5k+3\right)\left(5k+3\right)=1200:3:4\)

\(\Rightarrow\left(5k+3\right)^2=100\)

\(\Rightarrow\left[{}\begin{matrix}5k+3=10\\5k+3=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}k=\dfrac{7}{5}\\k=\dfrac{-13}{5}\end{matrix}\right.\)

+) 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 .................

MÌNH KO BIẾT ĐÚNG KO ĐÂU NHA

pt :15/(x-9)=20/(y-12) <=> 60/(4x-36)=60/(3y-36) : (Quy đồng mẫu) 
=> 4x=3y 
<=> x= 3y/4 
kết hợp với xy= 1200 => x=30 hoặc x=-30 =>y =+-40 
thế x hoặc y vào pt ban đàu ta có z= 80                            (pt là phân tích, mìh ko bít gõ phân số nên thông cảm :D)