Đặt x/5=k => x = 5k
y/3= k => y = 3k
Theo bài ra ta có : x . y = 60
Hay 5k . 3k = 60
<=> 15k2 = 60
<=> k2 = 4
<=> k = +4 hoặc k = -4
Vậy x = 20 hoặc x = -20
y = 12 hoặc y = -12
x/5 = y/3 = k
=> x = 5k; y = 3k
=> xy = 5k.3k = 15k2 = 60
=> k2 = 4
=> k = 2 hoặc k = -2
*k = 2 => x = 2.5 = 10; y = 2.3 = 6
*k = -2 => x = -2.5 = -10; y = -2.3 = -6
vậy_
\(\frac{x}{5}=\frac{y}{3}\Rightarrow\left(\frac{x}{5}\right)^2=\frac{x}{5}.\frac{y}{3}=\frac{x.y}{15}=\frac{60}{15}=4\)
\(\frac{x^2}{25}=4\)
\(x^2=100\)
\(\orbr{\begin{cases}x=-10\\x=10\end{cases}\Rightarrow}\orbr{\begin{cases}y=-6\\y=6\end{cases}}\)
Vậy \(\left(x,y\right)=\left\{\left(10,6\right);\left(-10,-6\right)\right\}\)