Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
=>x=2k; y=3k
xy=9
=>\(2k\cdot3k=9\)
=>\(k^2=\dfrac{9}{6}=\dfrac{3}{2}\)
TH1: \(k=\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\)
\(x=2\cdot\dfrac{\sqrt{6}}{2}=\sqrt{6};y=3\cdot\dfrac{\sqrt{6}}{2}=\dfrac{3\sqrt{6}}{2}\)
TH2: \(k=-\dfrac{\sqrt{6}}{2}\)
=>\(x=2\cdot\dfrac{-\sqrt{6}}{2}=-\sqrt{6};y=3\cdot\dfrac{-\sqrt{6}}{2}=-\dfrac{3\sqrt{6}}{2}\)