\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=>\frac{xyz}{12.9.5}=\frac{x^3}{12^3}=\frac{20}{540}\Rightarrow x^3=\frac{20.12^3}{540}=64\Rightarrow x=4\)

tự làm tiếp nhé 

\(\frac{xyz}{12.9.5}=\frac{y^3}{9^3}=\frac{20}{540}=>.......y=.......\)

\(\frac{xyz}{12.9.5}=\frac{z^3}{5^3}=\frac{20}{540}=>.....z=......\)

Đặt \(\frac{x}{12}\)=\(\frac{y}{9}\)=\(\frac{z}{5}\)=k

=>x=12k, y=9k, z=5k

Vì xyz=20=> 12k.9k.5k=20

              => 540k³=20

              => k³=\(\frac{1}{27}\)

              => k=\(\frac{1}{3}\)=> x=4, y=3,z=\(\frac{5}{3}\)

\(x:y:z=12:9:5\Rightarrow\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)

\(\Rightarrow x=12k;y=9k;z=5k\)

\(\Rightarrow xyz=12k.9k.5k=540k^3\)

\(\Rightarrow540k^3=20\Rightarrow k^3=\dfrac{1}{27}\Leftrightarrow k=\dfrac{1}{3}\)

\(\Leftrightarrow x=12.k=12.\dfrac{1}{3}=4\)

\(\Leftrightarrow y=9.k=9.\dfrac{1}{3}=3\)

\(\Leftrightarrow z=5.k=5.\dfrac{1}{3}=\dfrac{5}{3}\)

Vậy,....

x/12=y/9=z/5 = k => x = 12k ; y = 9k ; z = 5k

Thay vào ta được:

12k.9k.5k = 20

540k³  = 20

k³ = 1/27

Vậy k = 1/3

x = 1/3 . 12 = 4

y = 9.1/3 = 3

z = 1/3 . 5 = 5/3 

đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

\(\Rightarrow x=12k;y=9k;z=5k\)

Mà xyz = 20

\(\Rightarrow\)12k . 9k . 5k = 20

\(\Rightarrow\)540k³ = 20

\(\Rightarrow\)k³ = \(\frac{1}{27}\)

\(\Rightarrow\)k = ( -3 )

\(\Rightarrow\)x = -36 ; y = -27 ; z = -15

Ta có:

\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\Leftrightarrow x=12k;y=9k;z=5k\) và \(xyz=20\)

\(\Rightarrow12k.9k.5k=20\)

\(\Rightarrow540k^3=20\Leftrightarrow k=\sqrt[3]{20:540}=\frac{1}{3}\)

\(\hept{\begin{cases}x=12.\frac{1}{3}=4\\y=9.\frac{1}{3}=3\\z=5.\frac{1}{3}=\frac{5}{3}\end{cases}}\)

Vậy x = 4; y = 3 ; z = 5/3

Ta có: \(\left[\begin{array}{nghiempt}xyz=20\\\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}xyz=20\\x=12k\\y=9k\\z=5k\end{array}\right.\)

\(\Rightarrow xyz=12k.9k.5k=540k^3\)

\(\Rightarrow20=540k^3\)

\(\Rightarrow k^3=\frac{20}{540}=\frac{1}{27}\Rightarrow k^3=\left(\frac{1}{3}\right)^3\Rightarrow k=\frac{1}{3}\)

\(\Rightarrow x=12k=12.\frac{1}{3}=4\)

\(\Rightarrow y=9k=9.\frac{1}{3}=3\)

\(\Rightarrow z=5k=\frac{5.1}{3}=\frac{5}{3}\)

TA CÓ X/12=Y/9=Z/5 =>X=12K;Y=9K;Z=5K

MÀ XYZ=20=>12K.9K.5K=20 HAY 540\(K^3\)=20

=>\(K^3\)=20/540=1/27=>\(K^3\)=\(\left(\frac{1}{3}\right)^3\)=>K=1/3

TỪ X/12=1/3=>X=4

      Y/9=1/3=>Y=3

      Z/5=1/3=>Z=5/3

VẬY X=4;Y=3;Z=5/3

TICK ĐÚNG CHO MIK NHA

\(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}\)

Đặt : \(\dfrac{x}{12}=k\Leftrightarrow x=12k\)

\(\dfrac{y}{9}=k\Leftrightarrow y=9k.\)

\(\dfrac{z}{5}=k\Leftrightarrow z=5k\)

Theo đề bài ta có : \(12k.9k.5k=20\)

\(540.k^3=20\)

\(\Leftrightarrow k=\dfrac{1}{3}\)

Thay vào ta sẽ có :

\(x=4,y=3,z=\dfrac{5}{3}\)

Giải:

Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\)

Ta có: \(xyz=20\)

\(\Rightarrow540k^3=20\)

\(\Rightarrow k^3=\dfrac{1}{27}\)

\(\Rightarrow k=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{4}\\z=\dfrac{5}{12}\end{matrix}\right.\)

Vậy...

Vì  x: y : z = 12 : 9 : 5  nên \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)

Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

\(\Rightarrow\hept{\begin{cases}x=12k\\y=9k\\z=5k\end{cases}}\)

Thay vào ta có :

\(12k.9k.5k=20\)

\(540.k^3=20\)

\(k^3=\frac{1}{27}\)

\(k^3=\left(\frac{1}{3}\right)^3\)

\(k=\frac{1}{3}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{12}=\frac{1}{3}\\\frac{y}{9}=\frac{1}{3}\\\frac{z}{5}=\frac{1}{3}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=\frac{5}{3}\end{cases}}\)

x:y:z=12:9:5 ->\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)

Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)= k -> x = 12*k ;y = 9*k ;z = 5*k (1)

thay (1) vào xyz=20 ta được : 12k + 9k