Đặt \(\dfrac{x}{5}=\dfrac{y}{7}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
Lại có : \(xy=1715\)
\(\Leftrightarrow5k.7k=1715\)
\(\Leftrightarrow35k^2=1715\)
\(\Leftrightarrow k^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}k=7\\k=-7\end{matrix}\right.\)
+) \(k=7\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5.7=35\\y=7.7=49\end{matrix}\right.\)
+) \(k=-7\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5.\left(-7\right)=-35\\y=7.\left(-7\right)=-49\end{matrix}\right.\)
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 \(x=-35,y=-49\)
