Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(xy=10\)
\(\text{Ta được : }2k\cdot3k=10\\ \Leftrightarrow6k^2=10\\ \Leftrightarrow k^2=\dfrac{5}{3}\\ \Leftrightarrow k=\sqrt{\dfrac{5}{3}}\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\sqrt{\dfrac{5}{3}}\\y=3\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)
Vậy \(x=2\sqrt{\dfrac{5}{3}};y=3\sqrt{\dfrac{5}{3}}\)
\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow3x=2y\Rightarrow x=\dfrac{2y}{3}\) thay vào xy=10 ta có:
\(\dfrac{2y}{3}.y=10\Leftrightarrow2y^2=30\Leftrightarrow y^2=15\Leftrightarrow y=\sqrt{15}\)
\(\Rightarrow x=\dfrac{10}{\sqrt{15}}\)
