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

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x+y-z}{2\cdot4+3-2}=\dfrac{9}{9}=1\)

Do đó: x=4; y=3; z=2

\(\dfrac{3x}{2}=\dfrac{4y}{3}=\dfrac{6z}{5}\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{5}{6}}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{5}{6}}=\dfrac{x+y-z}{\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{5}{6}}=\dfrac{-21}{\dfrac{7}{12}}=-36\\ \Rightarrow\left\{{}\begin{matrix}x=-36\cdot\dfrac{2}{3}=-24\\y=-36\cdot\dfrac{3}{4}=-27\\z=-36\cdot\dfrac{5}{6}=-30\end{matrix}\right.\)

Ta có: 3x = 4y => \(\frac{x}{4}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{6}\)

        5y = 6z => \(\frac{y}{6}=\frac{z}{5}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có: 

       \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y-z}{8+6-5}=\frac{18}{9}=2\)

=> \(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{6}=2\\\frac{z}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.8=16\\y=2.6=12\\z=2.5=10\end{cases}}\)

Vậy ....

\(\frac{2}{3x}=\frac{3}{4y}=\frac{5}{6z}\Rightarrow\frac{2}{30.3x}=\frac{3}{30.4y}=\frac{5}{30.6z}\Leftrightarrow\frac{1}{45x}=\frac{1}{40y}=\frac{1}{36z}\Rightarrow45x=40y=36z\)

\(\Rightarrow x=\frac{9}{8}y;x=\frac{5}{4}z\Rightarrow x^2+y^2+z^2=x^2+\frac{64}{81}x^2+\frac{16}{25}x^2\)

\(=x^2\left(1+\frac{2896}{2025}\right)=724\text{ :)) đến đây thôi :))}\)

Ko bạn ơi , mik ko viết rõ nên bạn lầm á . Phải là \(\frac{2}{3}x=\frac{3}{4}y=\frac{5}{6}z\)và x²+ y²+ z²=724

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

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

3x=4y

=>x/4=y/3

=>x/8=y/6

5y=6z

=>y/6=z/5

=>x/8=y/6=z/5

Đặt x/8=y/6=z/5=k

=>x=8k; y=6k; z=5k

xyz=30

=>8k*6k*5k=30

=>240k^3=30

=>k^3=1/8

=>k=1/2

=>x=8*1/2=4; y=6*1/2=3; z=5*1/2=5/2

1)

a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )

5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)

\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)

b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)

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

\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)