Theo đề bài ta có :
\(\dfrac{x}{5}=\dfrac{y}{3}\) và \(x^2-y^2=4\)
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)
mà \(x^2-y^2=4\)
hay \(\left(5k\right)^2-\left(3k\right)^2=4\)
\(\Rightarrow25.k^2-9.k^2=4\)
\(\Rightarrow k^2.\left(25-9\right)=4\)
\(\Rightarrow k^2=4:\left(25-9\right)\)
\(\Rightarrow k^2=0,25\)
\(\Rightarrow k=\left(\pm0,5\right)\)
Với \(k=0,5\Rightarrow x=2,5;y=1,5\)
Với \(k=-0,5\Rightarrow x=-2,5;y=-1,5\)
Vậy...........
Ta có: 3x=5y và x2-y2=4
\(\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{x^2-y^2}{\dfrac{1}{9}-\dfrac{1}{25}}=\dfrac{4}{\dfrac{16}{225}}=\dfrac{225}{4}\)
*3x=\(\dfrac{225}{4}\)\(\Rightarrow x=\dfrac{225}{4}:3=\dfrac{75}{4}\)
*5y=\(\dfrac{225}{4}\Rightarrow y=\dfrac{225}{4}:5=\dfrac{45}{4}\)
