Ta có : \(x:y:z=12:9:5\Rightarrow\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)
Ta có : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
⇒ \(x=12k\)
\(y=9k\)
\(z=5k\)
Mà \(x.y.z=20\)
\(\Rightarrow12k.9k.5k=20\)
\(k^3.540=20\)
\(k^3=20:540\)
\(k^3=\frac{1}{27}\)
\(k^3=\left(\frac{1}{3}\right)^3\)
\(k=\frac{1}{3}\)
⇒ \(\frac{x}{12}=\frac{1}{3}\Rightarrow x=12.\frac{1}{3}=4\)
\(\frac{y}{9}=\frac{1}{3}\Rightarrow y=9.\frac{1}{3}=3\)
\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=5.\frac{1}{3}=\frac{5}{3}\)
Vậy \(x=4;y=3;z=\frac{5}{3}\)
Ta có: \(x:y:z=12:9:5.\)
=> \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\) và \(x.y.z=20\)
Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=12k\\y=9k\\z=5k\end{matrix}\right.\)
Có: \(x.y.z=20\)
=> \(12k.9k.5k=20\)
=> \(540k^3=20\)
=> \(k^3=20:540\)
=> \(k^3=\frac{1}{27}\)
=> \(k=\frac{1}{3}.\)
+) Với \(k=\frac{1}{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}.12=4\\y=\frac{1}{3}.9=3\\z=\frac{1}{3}.5=\frac{5}{3}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(4;3;\frac{5}{3}\right).\)
Chúc bạn học tốt!
