Ta có \(x:y=2:3\)
=> \(\frac{x}{y}=\frac{2}{3}.\)
=> \(\frac{x}{2}=\frac{y}{3}\) và \(x.y=54.\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k=\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
\(x.y=54\)
=> \(2k.3k=54\)
=> \(6k^2=54\)
=> \(k^2=54:6\)
=> \(k^2=9\)
=> \(k=\pm3\)
TH1: \(k=3\)
=> \(\left\{{}\begin{matrix}x=3.2=6\\y=3.3=9\end{matrix}\right.\)
TH2: \(k=-3\)
=> \(\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).3=-9\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;9\right),\left(-6;-9\right).\)
Chúc bạn học tốt!

