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

Chia từng trường hợp tìm y, z.

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)

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 \(\left(x;y;z\right)\in\left\{\left(30;40;80\right);\left(-30;-40;-80\right)\right\}\)

Ta có : 15/(x-9)= 20/(y-12)

<=> 15(y-12) = 20(x-9)

<=> 15y - 180 = 20x - 180

<=> 3y            = 4x

<=> y              = 4/3x

Do xy = 1200 

=> 4/3. x^2 = 1200

=>  x^2  = 1200 : 4/3

=> x^2 = 900

<=> x = 30

<=> y = 40

<=> 5/7 = 40/(z-24)

<=> 80 = z 

=> x=30 ; y=40 ; z=80

Mọi người giải giúp mk với nha :))

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

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

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

\(\Rightarrow\frac{3}{4}=\frac{x-9}{y-12}=\frac{9}{12}=\frac{x-9}{y-12}=\frac{x-9+9}{y-12+12}\)\(=\frac{x}{y}=\frac{xy}{y^2}=\frac{x^2}{xy}\)

Từ \(\frac{3}{4}=\frac{xy}{y^2}\Rightarrow\frac{3}{4}=\frac{1200}{y^2}\Rightarrow y^2=1200\cdot\frac{4}{3}=20^2\Rightarrow y=\pm40\)

• Nếu y=40 => x= 1200: 40 = 30 

Mà \(\frac{15}{x-9}=\frac{40}{z-24}\Rightarrow z=80\)

• Nếu y = -40 => x = 1200:(-40) = - 30 

Mà \(\frac{15}{x-9}=\frac{40}{z-24}\Rightarrow z=-80\)

Vây (x , y , z ) = ( 30, 40, 80); ( - 30; -40; -80)