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

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

Chia từng trường hợp tìm y, z.

x=30.y=40.z=80

Mình nghĩ là bạn chép sai đề bài chỗ 20/x-12
\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\)
\(\frac{15}{x-9}=\frac{20}{y-12}\)
\(\Rightarrow\frac{y-12}{x-9}=\frac{20}{15}=\frac{4}{3}=\frac{12}{9}\)
\(\frac{y-12}{x-9}=\frac{12}{9}=\frac{y-12+12}{x-9+9}=\frac{y}{x}=\frac{4}{3} \)
\(\frac{y}{x}=\frac{4}{3}=>\frac{y}{4}=\frac{x}{3}=k\)
\(\Rightarrow x=3k,y=4k\)
\(xy=4k.3k\)
\(\Rightarrow12k^2=1200\)
         \(k^2=1200:12=100=10^2=-10^2\)
       \(k=10hoac=-10\)
Nếu k = 10 thì
x=3.10=30
y=4.10=40
Nếu k= -10 thì
x=.........
y=........
Ta có:
\(\frac{40}{z-24}=\frac{15}{x-9}=\frac{15}{30-9}=\frac{5}{7}\)
\(\frac{40}{z-24}=\frac{5}{7}=\frac{40}{56}\)
=> z-24=56
     z    =56+24=80
Nếu x= -30
Ta có :..............
Phần còn lại bạn tính z= - 80
        


 

CHAO BAN MINH SANG MOI KET BAN DAY TICH CHO MINH NHA

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

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

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

\(=\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)

\(\Rightarrow\frac{x^2}{15^2}=\frac{y^2}{20^2}=\frac{z^2}{40^2}=\frac{xy}{15.20}=\frac{1200}{300}=4=2^2\)

\(\Rightarrow\begin{cases}x^2=2^2.15^2=30^2\\y^2=2^2.20^2=40^2\\z^2=2^2.40^2=80^2\end{cases}\)\(\Rightarrow\begin{cases}x\in\left\{30;-30\right\}\\y\in\left\{40;-40\right\}\\z\in\left\{80;-80\right\}\end{cases}\)

Vậy giá trị (x;y;z) tương ứng thỏa mãn là: (30;40;80) ; (-30;-40;-80)

Ta có:

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

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

\(\Rightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=15k\\y=20k\\z=40k\end{matrix}\right.\)

\(\Rightarrow xy=15k20k=300k^2=1200\)

\(\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1:\(k=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=15k=30\\y=20k=40\\z=40k=80\end{matrix}\right.\)

TH2:\(k=-2\)

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

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

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

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

=> \(\frac{x}{3}\) = \(\frac{y}{4}\)

vậy x = 30, y = 40 ,  z = 80

X=30 Y=40 Z=80