Ta có: \(x\div y\div z=2\div3\div5\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\inℝ\right)\)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Mà xyz = 30 \(\Rightarrow30k^3=30\Rightarrow k=1\)

\(\Rightarrow\hept{\begin{cases}x=2\\y=3\\z=5\end{cases}}\)

Theo bài ra ta có : 

x : y : z = 2 : 3 : 5 và x.y.z = 30

=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Gọi\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)=>\(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Thay x=2k; y=3k; z=5k vào x.y.z = 30 ta có :

       2k.3k.5k=30

<=> k³.30=30

=> k³=1

=> k=1

=>\(\hept{\begin{cases}x=2.1=2\\y=3.1=3\\z=5.1=5\end{cases}}\)

Tỉ lệ : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Đặt \(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\Rightarrow xyz=30\)

Ta có : \(2k.3k.5k=30\Leftrightarrow30k^3=30\Leftrightarrow k=1\)

Vậy \(x=2;y=3;z=5\)

a.

Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)

Thế vào \(2x+y-z=81\)

\(\Rightarrow2.5k+3k-4k=81\)

\(\Rightarrow9k=81\)

\(\Rightarrow k=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)

b.

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)

Thế vào \(5x-y+3z=124\)

\(\Rightarrow5.3k-5k+3.2k=124\)

\(\Rightarrow16k=124\)

\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)

c.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Thế vào \(xyz=810\)

\(\Rightarrow2k.3k.5k=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)

d.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Thế vào \(x^2y^2z^2=288^2\)

\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)

\(\Rightarrow\left(k^2\right)^3=64\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)

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...

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=k\)

\(\Rightarrow xyz=5.2.\left(-3\right).k=-30k=240\Rightarrow k=-8\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-8\right).5=-40\\y=\left(-8\right).2=-16\\z=\left(-8\right).\left(-3\right)=24\end{matrix}\right.\)