Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)

Và \(480=60k^3\Rightarrow k=2\) suy ra x = 6; y =8, z =10 

Vậy....

Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\) =>

=> \(3k.4k.5k=x.y.z\)

=> \(\left(3.4.5\right).k^3=480\)

=> \(60.k^3=480\)

=> \(k^3=\frac{480}{60}=8\)

=> \(k=2\)

Từ đó suy ra: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=2\)

=> \(x=3.2=6\)

=> \(y=4.2=8\)

=> \(z=5.2=10\)

Đặt k = \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

\(\Rightarrow x=3k,y=4k,z=5k\)

Từ x.y.z = 480, ta có:

3k.4k.5k = 480

\(\Rightarrow60k^3\) = 480

\(\Rightarrow k^3=8\)

\(\Rightarrow k=2\)

Với k = 2 \(\Rightarrow\) x = 6, y = 8, z = 10

Lời giải :

a) Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=k\)

\(\Leftrightarrow\hept{\begin{cases}x=5k\\y=4k\\z=2k\end{cases}}\)

Ta có : \(xyz=40k^3=240\)

\(\Leftrightarrow k^3=6\)

\(\Leftrightarrow k=\sqrt[3]{6}\)

\(\Leftrightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=\sqrt[3]{6}\)

\(\Leftrightarrow\hept{\begin{cases}x=5\sqrt[3]{6}\\y=4\sqrt[3]{6}\\z=2\sqrt[3]{6}\end{cases}}\)

Vậy....

b) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{9}=\frac{y}{6}\)

Ta cũng có \(\frac{y}{3}=\frac{z}{2}\Leftrightarrow\frac{y}{6}=\frac{z}{4}\)

Khi đó : \(\frac{x}{9}=\frac{y}{6}=\frac{z}{4}=\frac{x-y+z}{9-6+4}=\frac{21}{7}=3\)

\(\Leftrightarrow\hept{\begin{cases}x=27\\y=18\\z=12\end{cases}}\)

Vậy...