Đặt \(k=\frac{x}{8}=\frac{y}{3}=\frac{z}{10}\)

Ta có: \(x=8k;y=3k;z=10k\)  (*)

Thay vào đẳng thức \(xy+yz+zx=206\) ta được:

  \(8k.3k+3k.10k+10k.8k=206\)

\(\Leftrightarrow24k^2+30k^2+80k^2=206\)

\(\Leftrightarrow24k^2+30k^2+80k^2=206\)

\(\Rightarrow k=\pm\sqrt{\frac{103}{67}}\)

Thay k vào (*) tính được x, y, z

\(\Rightarrow\left(x.y.z\right)^2=\frac{-2}{5}.\frac{3}{4}.\frac{-3}{10}\)

\(\Rightarrow\left(x.y.z\right)^2=\frac{18}{200}=\frac{9}{100}\)

\(\Rightarrow x.y.z=\frac{3}{10}\)

\(\Rightarrow z=\frac{3}{-4}\)

\(\Rightarrow x=\frac{2}{5}\)

\(\Rightarrow y=-1\)

\(\left(xy\right):\left(yz\right)=\frac{2}{3}:0,6\Rightarrow\frac{x}{z}=\frac{10}{9}\)=> \(x=\frac{10}{9}z\Rightarrow\frac{10}{9}z.z=0,625\Rightarrow z^2=\frac{9}{16}\Rightarrow z=\pm\frac{3}{4}\)

\(\left(yz\right):\left(zx\right)=0,6:0,625\Rightarrow\frac{y}{x}=\frac{24}{25}\)

Với z=3/4 => x, y

Với z=-3/4 => x,y

Câu b làm tương tự nhé :)

a,\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)=3  

x= 1

y=1

z=1

x=1

y=1

z=1

\(x\cdot y\cdot y\cdot z\cdot z\cdot x=\frac{2}{3}\cdot0,6\cdot0,625\)

\(x^2\cdot y^2\cdot z^2=\frac{1}{4}\)

\(\left(x\cdot y\cdot z\right)^2=\frac{1}{4}\)

\(x\cdot y\cdot z=\frac{1}{2}\) hoặc \(x\cdot y\cdot z=-\frac{1}{2}\)

Nếu \(x\cdot y\cdot z=\frac{1}{2}\) thì

\(x=\frac{1}{2}:\left(y\cdot z\right)=\frac{1}{2}:0,6=\frac{5}{6}\)

\(y=\frac{1}{2}:\left(z\cdot x\right)=\frac{1}{2}:0,625=\frac{4}{5}\)

\(z=\frac{1}{2}:\left(x\cdot y\right)=\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\)

Nếu \(x\cdot y\cdot z=-\frac{1}{2}\) thì

\(x=\left(-\frac{1}{2}\right):\left(y\cdot z\right)=\left(-\frac{1}{2}\right):0,6=-\frac{5}{6}\)

\(y=\left(-\frac{1}{2}\right):\left(z\cdot x\right)=\left(-\frac{1}{2}\right):0,625=-\frac{4}{5}\)

\(z=\left(-\frac{1}{2}\right):\left(x\cdot y\right)=\left(-\frac{1}{2}\right):\frac{2}{3}=-\frac{3}{4}\)

nhân từng vế 3 đẳng thức ta đc:

xy.yz.zx=2/3.0,6.0,625

=>(xyz)^2=1/4=(+1/2)^2

+)với xyz=1/2

cùng xy=>z=1/2:2/3=3/4

cùng yz=...=>...

cùng zx=..=>...

+)với xyz=-1/2 (làm tương tự)

KL:có 2 cặp bộ ba (x;y;z)  thỏa mãn là....