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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!

\(a.\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\&xy=1200\)

\(\Leftrightarrow\dfrac{15}{20}=\dfrac{x-9}{y-12}\Leftrightarrow\dfrac{3}{4}=\dfrac{x-9}{y-12}\)

\(\Rightarrow\dfrac{9}{12}=\dfrac{x-9}{y-12}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{9}{12}=\dfrac{x-9}{y-12}=\dfrac{x-9+9}{y-12+12}=\dfrac{x}{y}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{xy}{y^2}=\dfrac{x^2}{xy}\)

Từ \(\dfrac{3}{4}=\dfrac{xy}{y^{^2}}\Rightarrow\dfrac{3}{4}=\dfrac{1200}{y^2}\Rightarrow y^2=1200.\dfrac{4}{3}=1600\)

\(\Rightarrow y=\sqrt{1600}=\pm40\)

+ TH1: \(y=40\Rightarrow x=30\)

\(\dfrac{15}{x-9}=\dfrac{40}{z-24}\Rightarrow z=80\) (tự giải pt)

+ TH2: \(y=-40\Rightarrow x=-30\)

\(\dfrac{15}{x-9}=\dfrac{40}{z-4}\Rightarrow z=-80\) (tự giải pt)

Vậy, các cặp \(\left(x;y;z\right)\) thỏa mãn là \(\left(30;40;80\right)\&\left(-30;-40;-80\right)\)

\(b.15x=-10y=6z\&xyz=30000\)

\(\Rightarrow\left\{{}\begin{matrix}15x=-10y\\-10y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-10}=\dfrac{y}{15}\\\dfrac{y}{6}=\dfrac{z}{-10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-20}=\dfrac{y}{30}\\\dfrac{y}{30}=\dfrac{z}{-50}\end{matrix}\right.\Rightarrow\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}\)

Đặt \(\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}=k\Rightarrow x=-20k;y=30k;z=-50k\)

\(\Rightarrow xyz=30000\Rightarrow-20k.30k.\left(-50k\right)=30000\Rightarrow30000k^3=30000\)

\(\Rightarrow k^3=1\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=30\\z=-50\end{matrix}\right.\)

⊥

Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)

\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)

kb hộ

em chịu 

làm bài hộ mk nhé

Nhìu kết quả lắm vì có rất nhiều số nhân với nhau = 1200 nhé

khó quá

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 : 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 :))

minh cung eo hieu j sorry