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

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/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 \(x=4;y=3;z=\frac{5}{3}\)

1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2\cdot9+4\cdot6}=\frac{13}{26}=\frac12\)

=>\(\begin{cases}x=20\cdot\frac12=10\\ y=9\cdot\frac12=\frac92\\ z=6\cdot\frac12=3\end{cases}\)

2: \(\frac{x}{3}=\frac{y}{4}\)

=>\(\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{7}\)

=>\(\frac{y}{20}=\frac{z}{28}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

mà 2x+3y-z=186

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y-z}{2\cdot15+3\cdot20-28}=\frac{186}{62}=3\)

=>\(\begin{cases}x=3\cdot15=45\\ y=3\cdot20=60\\ z=3\cdot28=84\end{cases}\)

3: \(\frac{x}{2}=\frac{2y}{5}=\frac{4z}{7}\)

=>\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}\)

mà 3x+5y+7z=123

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}=\frac{3x+5y+7z}{3\cdot2+5\cdot2,5+7\cdot1,75}=\frac{123}{30,75}=4\)

=>\(\begin{cases}x=4\cdot2=8\\ y=4\cdot2,5=10\\ z=4\cdot1,75=7\end{cases}\)

4: \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)

=>\(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}\)

Đặt \(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}=k\)

=>\(x=2k;y=\frac32k;z=\frac43k\)

xyz=-108

=>\(2k\cdot\frac32k\cdot\frac43k=-108\)

=>\(4k^3=-108\)

=>\(k^3=-27\)

=>k=-3

=>\(\begin{cases}x=2\cdot\left(-3\right)=-6\\ y=\frac32\cdot\left(-3\right)=-\frac92\\ z=\frac43\cdot\left(-3\right)=-4\end{cases}\)

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