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

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

\(\Rightarrow\frac{x}{12}=K\Rightarrow x=12K\)

\(\Rightarrow\frac{y}{9}=K\Rightarrow y=9K\)

\(\Rightarrow\frac{z}{5}=K\Rightarrow z=5K\)

Mà \(12K\cdot9K\cdot5K=20\)

\(\Rightarrow\left(12\cdot9\cdot5\right)\cdot K=20\)

\(\Rightarrow540k=20\)

\(\Rightarrow K=20:540\)

\(\Rightarrow K=+-\frac{1}{27}\)

KHi đó \(\frac{x}{12}=\frac{1}{27}\Rightarrow x=\frac{4}{9}\)

\(\frac{y}{9}=\frac{1}{27}\Rightarrow y=\frac{1}{3}\)

\(\frac{z}{5}=\frac{1}{27}\Rightarrow z=\frac{5}{27}\)

Còn lại bn lm tương tự ha

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

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

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

-->12.9.5.k³=20

-->540.k³=20

-->k³=\(\frac{1}{27}\)-->k=\(\frac{1}{3}\)

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

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

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

Vậy x=4;y=3;z=\(\frac{5}{3}\)

sai rồi  bn ơi.

là nhân chứ có phải là cộng đâu mà phân phối vậy?

Câu a) sai đề nhé bạn.

b) Ta có:

\(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\) và \(2x+5y-2z=100\)

\(\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}\) và \(2x+5y-2z=100\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{2x+5y-2z}{2.7+5.20-2.32}=\frac{100}{50}=2\)

\(\hept{\begin{cases}\frac{x}{7}=2\Rightarrow x=2.7=14\\\frac{y}{20}=2\Rightarrow y=2.20=40\\\frac{z}{32}=2\Rightarrow z=2.32=64\end{cases}}\)

Vậy \(x=14;y=40;z=64\)

\(TC:\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(KĐ:\) \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot9=-27\\y=-3\cdot7=-21\\z=-3\cdot3=-9\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

AD TC DTSBN ta có 

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\dfrac{x}{9}=-3\Rightarrow x=-27\)

\(\dfrac{y}{7}=-3\Rightarrow y=-21\)

\(\dfrac{z}{3}=-3\Rightarrow z=-9\)

Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{7}\)(1)

Ta có: \(\dfrac{y}{z}=\dfrac{7}{3}\)

\(\Leftrightarrow\dfrac{y}{7}=\dfrac{z}{3}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

mà x-y+z=-15

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

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{9}=-3\\\dfrac{y}{7}=-3\\\dfrac{z}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

Vậy: (x,y,z)=(-27;-21;-9)

1) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}=\frac{2x-4y+3z}{2.2-4.3+3.7}=\frac{-39}{13}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.2=-6\\y=-3.3=-9\\z=-3.7=-21\end{cases}}\)

2) \(9x=10y\Leftrightarrow\frac{x}{10}=\frac{y}{9},4y=3z\Leftrightarrow\frac{y}{9}=\frac{z}{12}\)

suy ra \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Leftrightarrow\hept{\begin{cases}x=6.10=60\\y=6.9=54\\z=6.12=72\end{cases}}\)

3) \(3x=4y=6z\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x-y+z}{4-3+2}=\frac{-9}{3}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.4=-12\\y=-3.3=-9\\z=-3.2=-6\end{cases}}\)

\(\dfrac{x}{y+z-3}=\dfrac{y}{x+z}=\dfrac{z}{x+y+3}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=\dfrac{1}{4044\left(x+y+z\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}y+z-3=2x\\x+z=2y\\x+y+3=2z\end{matrix}\right.\) và \(4044\left(x+y+z\right)=2\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=3x+3\\x+y+z=3y\\x+y+z=3z-3\end{matrix}\right.\\ \Rightarrow3x+3=3y=3z-3\\ \Rightarrow x+1=y=z-1\)

\(\left\{{}\begin{matrix}x=y-1\\z=y+1\end{matrix}\right.\)

Lại có \(4044\left(x+y+z\right)=2\)

\(\Rightarrow4044\left(y-1+y+y+1\right)=2\\ \Rightarrow4044\cdot3y=2\\ \Rightarrow y=\dfrac{1}{674}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{673}{674}\\z=\dfrac{675}{674}\end{matrix}\right.\)