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\(\frac{15}{x-9}=\frac{20}{y-12}\Leftrightarrow15\left(y-12\right)=20\left(x-9\right)\Leftrightarrow15y-180=20x-180\Leftrightarrow15y=20x\Leftrightarrow\frac{y}{20}=\frac{x}{15}\Leftrightarrow\frac{xy}{20}=\frac{x^2}{15}\Leftrightarrow x^2=\frac{15.1200}{20}=900\Leftrightarrow\orbr{\begin{cases}x=30\\x=-30\end{cases}}\)
Chia từng trường hợp tìm y, z.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\left(x;y;z\right)\in\left\{\left(30;40;80\right);\left(-30;-40;-80\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{24}{z-24}\Rightarrow\)\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)
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\(\hept{\begin{cases}x=15.\frac{-13}{5}+9=-30\\y=20.\frac{-13}{5}+12=-40\\z=40.\frac{-13}{5}+24=-80\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{y-12}\)
\(\Rightarrow\frac{3}{4}=\frac{x-9}{y-12}=\frac{9}{12}=\frac{x-9}{y-12}=\frac{x-9+9}{y-12+12}\)\(=\frac{x}{y}=\frac{xy}{y^2}=\frac{x^2}{xy}\)
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)