co 4 so do la 21

42

43

84

vay co 4 so thao man yeu cau de bai

không đúng đề bài nha

Đặt \(\dfrac{x}{5}=\dfrac{y}{7}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)

Ta có:

\(x.y=1715\)

\(\Rightarrow5k.7k=1715\)

\(\Rightarrow35.k^2=1715\)

\(\Rightarrow k^2=1715:35\)

\(\Rightarrow k^2=49\)

\(\Rightarrow k=7\) hoặc \(k=-7\)

Thay \(k=7\) ta có:

\(\left\{{}\begin{matrix}x=5k=5.7=35\\y=7k=7.7=49\end{matrix}\right.\)

Thay \(k=-7\) ta có:

\(\left\{{}\begin{matrix}x=5k=5.\left(-7\right)=-35\\y=7k=7.\left(-7\right)=-49\end{matrix}\right.\)

Vậy \(x=35;y=49\)

hoặc \(x=-35,y=-49\)