Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(3k\right)^2-\left(5k\right)^2=9k^2-25k^2=-16k^2=-4\)
\(\Rightarrow k^2=\dfrac{1}{4}\Rightarrow k=\pm\dfrac{1}{2}\)
\(\circledast k=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
\(\circledast k=-\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-\dfrac{5}{2}\end{matrix}\right.\)
\(Từ\dfrac{x}{3}=\dfrac{y}{5}=>\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{5}\right)^2=\dfrac{x^2}{9}=\dfrac{y^2}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\dfrac{x^2}{9}=\dfrac{y^2}{25}=\dfrac{x^2-y^2}{9-25}=\dfrac{-4}{-16}=\dfrac{1}{4}\)
=>\(x^2=\dfrac{1}{4}\cdot9=\dfrac{9}{4}=\left(+-\dfrac{3}{2}\right)^2\)
=>\(x=+-\dfrac{3}{2}\)
=>\(y^2=\dfrac{1}{4}\cdot25=\dfrac{25}{4}=\left(+-\dfrac{5}{2}\right)^2\)
=>\(y=+-\dfrac{5}{2}\)
Với \(x=\dfrac{3}{2}\) thì \(y=\dfrac{5}{2}\)
Với \(x=-\dfrac{3}{2}\) thì \(y=-\dfrac{5}{2}\)
