\(12x=-15y=10z\)

\(\Rightarrow\dfrac{12x}{60}=\dfrac{-15y}{60}=\dfrac{10z}{60}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}\)

Đặt: \(\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}=k\)

\(\Rightarrow x=5k;y=-4k;z=6k\)

Ta có: \(xyz=120\)

\(\Rightarrow5k\cdot-4k\cdot6k=120\)

\(\Rightarrow-120k^3=120\)

\(\Rightarrow k^3=-1\)

\(\Rightarrow k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-1\\\dfrac{y}{-4}=-1\\\dfrac{z}{6}=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-1\cdot5=-5\\y=-4\cdot-1=4\\z=-1\cdot6=-6\end{matrix}\right.\)

Vậy: ... 

\(\frac{12x-5y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=0\)\(=0\)

=> 12x - 15y =0

=> 12x = 15y

=> \(\frac{x}{15}=\frac{y}{12}\)

=> \(\frac{x}{60}=\frac{y}{48}\)

20z - 12x = 0 

=> 20z = 12x 

=> \(\frac{x}{20}=\frac{z}{12}\)

=> \(\frac{x}{60}=\frac{z}{36}\)

=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)

=> x = 1 . 60 : 3 = 20

y = 1 . 48 : 3 = 16

z = 1 . 36 : 3 = 12

Ta có:

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

Đề bài gì lạ vậy

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

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{27}=0\)

\(\Rightarrow\hept{\begin{cases}12x-15y=0\\15y-20z=0\end{cases}\Rightarrow\hept{\begin{cases}12x=15y\\15y=20z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{15}=\frac{y}{12}\\\frac{y}{20}=\frac{z}{15}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{75}=\frac{y}{60}\\\frac{y}{60}=\frac{z}{45}\end{cases}\Rightarrow}\frac{x}{75}=\frac{y}{60}=\frac{z}{45}}\)

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

\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)

=> x = 75.4 : 15 = 20 ; 

     y = 60.4 : 15 = 16

     z = 45.4 : 15 = 12

Vậy x = 20 ; y = 16 ; z = 12