Vì \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{y}{15}=\dfrac{x}{10}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Đặt \(\dfrac{x}{10}=\dfrac{z}{12}=\dfrac{y}{15}=k\)
\(\Rightarrow x=10k;z=12k\); \(y=15k\)
Thay vào đề bài ta được:
\(10k.12k=1080\)
\(\Rightarrow120k^2=1080\)
\(\Rightarrow k^2=3^2\)
\(\Rightarrow k=3\)
Khi đó \(\left[{}\begin{matrix}x=10.3=30\\z=12.3=36\\y=15.3=45\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=10.3=30\\z=12.3=36\\y=15.3=45\end{matrix}\right.\)
Cho mk sửa lại:
\(k^2=3^2\Rightarrow k=3;k=-3\)
_ k = 3 thì \(\left[{}\begin{matrix}x=10.\left(-3\right)=-30\\y=15.\left(-3\right)=-45\\z=12.\left(-3\right)=-36\end{matrix}\right.\)
_ k = -3 (mk xét kia rồi)
Vậy \(\left[{}\begin{matrix}x=\pm30\\y=\pm45\\z=\pm36\end{matrix}\right.\)
Cách khác:
\(\left\{\begin{matrix}\frac{x}{2}=\frac{y}{3}\Leftrightarrow x=\frac{2}{3}y\\\frac{y}{5}=\frac{z}{4}\Leftrightarrow z=\frac{4}{5}y\end{matrix}\right.\)\(\Rightarrow x.z=\frac{2}{3}y.\frac{4}{5}y\)
\(\Rightarrow\frac{8}{15}y^2=1080\)
\(\Rightarrow y^2=1080\div\frac{8}{15}\)
\(\Rightarrow y^2=2025\)
\(\Rightarrow\left\{\begin{matrix}y=\pm45\\x=\pm30\\z=\pm36\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) lần lượt là \(\left(30;45;36\right);\left(-30;-45;-36\right)\)