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 

\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)

Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)

\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186

Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)

Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)

\(c,x:y:z=12:9:5\)và xyz = 20

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

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

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

\(\Leftrightarrow xyz=12k\cdot9k\cdot5k=540k^3\)

\(\Leftrightarrow xyz=540k^3\)

\(\Leftrightarrow540k^3=xyz\)

\(\Leftrightarrow540k^3=20\)

\(\Leftrightarrow k^3=\frac{20}{540}=\frac{1}{27}\Leftrightarrow k=\frac{1}{3}\)

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

2) Đề thiếu rồi bạn.

3)

Ta có:

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

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

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

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

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

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

Với \(k=\frac{1}{3}.\)

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

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

Chúc bạn học tốt!

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

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

\(xyz=20\)

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

\(\Rightarrow540k^3=20\)

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

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

Khi   \(k=\frac{1}{3}\)

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

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

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

Vậy x = ..... ; y = ............ ; z = .............

a) ĐẶT \(\frac{x}{5}=\frac{y}{2}=k;\frac{x}{5}=k\Rightarrow x=5k;\frac{y}{2}=k\Rightarrow y=2k\)

ta có \(x.y=160\)

 thay\(5k.2k=160\)

\(k^2.10=160\)

\(k^2=16\)

\(\Rightarrow k=\pm4\)

do đó

 \(\frac{x}{5}=\pm4\Rightarrow\hept{\begin{cases}\frac{x}{5}=4\\\frac{x}{5}=-4\end{cases}\Leftrightarrow\hept{\begin{cases}x=5.4=20\\x=5.\left(-4\right)=-20\end{cases}}}\)

\(\frac{y}{2}=\pm4\Rightarrow\hept{\begin{cases}\frac{y}{2}=4\\\frac{y}{2}=-4\end{cases}\Leftrightarrow\hept{\begin{cases}y=2.4=8\\y=2.\left(-4\right)=-8\end{cases}}}\)

vậy các x,y thỏa mãn là \(\left\{x=20;y=8\right\}\left\{x=-20;y=-8\right\}\)

a) X*Y=160

=>X=160/Y (1)

X/5 =Y/2

=> 2x=5y(tính chất tỉ lệ thức)

=>x=5Y/2 (2)

(1),(2)=> 160/y = 5y/2

=> y=8

b) Ta có : \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{6}\)

\(\Rightarrow\frac{3x-3}{6}=\frac{4y-12}{16}=\frac{5z-25}{30}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{6}=\frac{3x-3}{6}=\frac{4y-12}{16}=\frac{5z-25}{30}=\frac{5z-25-3x+3-4y+12}{30-6-16}\)                                                                                                                                  \(=\frac{\left(5z-3x-4y\right)-10}{8}=\frac{50-10}{8}=\frac{40}{8}=5\)

\(\Rightarrow\hept{\begin{cases}x-1=10\\x-3=20\\x-5=30\end{cases}\Rightarrow\hept{\begin{cases}x=11\\y=23\\z=35\end{cases}}}\)

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

Khi đó xyz = 20

<=> 12k.9k.5k = 20

=> 540k³ = 20

=> \(k^3=\frac{1}{27}\)

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

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

a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)

8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)

=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)

=> x = 24,y = 15,z = 6

b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)

\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)

=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)

=> x = -165 , y = -20 , z = -25

c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k

=> xyz = 12k . 9k . 5k

=> xyz = 540k³

=> 540k³ =20

=> k³ = 20/540

=> k³ = 1/27

=> k = 1/3

Do đó : x= 4 , y = 3 , z = 5/3