Ta có: \(\dfrac{x}{2}=\dfrac{2y}{3}\) \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\)
Đặt \(\dfrac{x}{4}=\dfrac{y}{3}=k\)
\(\Rightarrow x=4k;y=3k\)
Thay vào \(xy=27\) có:
\(4k.3k=27\)
\(\Rightarrow12k^2=27\)
\(\Rightarrow k^2=\dfrac{9}{4}\Rightarrow k=\dfrac{9}{4}\)
Khi \(k=\dfrac{9}{4}\) thì:
\(x=4.\dfrac{9}{4}=9\)
\(y=3.\dfrac{9}{4}=\dfrac{27}{4}\)
Lúc này \(A=9^2+\left(\dfrac{27}{4}\right)^2=\dfrac{2025}{16}\)
Vậy \(A=\dfrac{2025}{16}.\)
Theo bài ra ta có: \(\dfrac{x}{2}=\dfrac{2y}{3}\)
Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}\) = k ( k \(\ne\) 0 )
\(\Rightarrow\dfrac{x.2y}{2.3}=\dfrac{2xy}{6}=\dfrac{2.27}{6}=\dfrac{54}{6}=9\) =\(k^2\)
\(\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3.2;y=3.3:2\\x=-3.2;y=-3.3:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6;y=4,5\\x=-6;y=-4,5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=20,25\end{matrix}\right.\)
\(\Rightarrow x^2+y^2=36+20,25\)
\(\Rightarrow A=56,25\)
