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

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 x/12 = y/9 = z/5 = k ta có:

x = 12k

y = 9k

z = 5k

=> x.y.z = 12k.9k.5k 

=> k^3.540=20

=> k^3 = 1/27

=> k^3= (1/3)^3

=> k = 1/3

x/12=1/3 => x=4

y/9= 1/3 => y=3

z/5=1/3 =. z=5/3

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

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

\(\Rightarrow xyz=12k.9k.5k=540k^3=20\)

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

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

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

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

\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=\frac{1}{3}.5=\frac{5}{3}\)

Vậy \(x=4;y=3;z=\frac{5}{3}\)

Hình như đề sai phải bạn 

Ta có : \(x:y:z=12:9:5\Rightarrow\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)

Ta có : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

⇒ \(x=12k\)

\(y=9k\)

\(z=5k\)

Mà \(x.y.z=20\)

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

\(k^3.540=20\)

\(k^3=20:540\)

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

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

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

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

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

\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=5.\frac{1}{3}=\frac{5}{3}\)

Vậy \(x=4;y=3;z=\frac{5}{3}\)

Ta có: \(x:y:z=12:9:5.\)

=> \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\) và \(x.y.z=20\)

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

Có: \(x.y.z=20\)

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

=> \(540k^3=20\)

=> \(k^3=20:540\)

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

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

+) Với \(k=\frac{1}{3}.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}.12=4\\y=\frac{1}{3}.9=3\\z=\frac{1}{3}.5=\frac{5}{3}\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(4;3;\frac{5}{3}\right).\)

Chúc bạn học tốt!

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