Ta có: 4x = 7y
<=> \(\dfrac{4}{y}=\dfrac{7}{x}\)
<=> \(\dfrac{16}{y^2}=\dfrac{49}{x^2}\)
=> \(\dfrac{16+49}{x^2+y^2}=\dfrac{65}{260}=\dfrac{1}{4}\)
=> \(\left\{{}\begin{matrix}x=28\\y=16\end{matrix}\right.\)
Lời giải:
$4x=7y\Rightarrow \frac{x}{7}=\frac{y}{4}=k$
$\Rightarrow x=7k; y=4k$
Ta có: $x^2+y^2=260$
$\Leftrightarrow (7k)^2+(4k)^2=260$
$\Leftrightarrow 65k^2=260$
$\Leftrightarrow k^2=4$
$\Rightarrow k=\pm 2$
Nếu $k=2$ thì $x=7k=14; y=4k=8$. Ta có $(x,y)=(14,8)$
Nếu $k=-2$ thì $x=7k=-14; y=4k=-8$. Ta có $(x,y)=(-14,-8)$