Có: \(\frac{3}{x}=\frac{y}{7}=\frac{x}{3}=\frac{y}{7}\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Thay \(x=3k;y=7k\) vào \(x.y=84\), ta có:
\(3k.7k=84\\ \Leftrightarrow21k^2=84\\ \Leftrightarrow k^2=4\\ \Leftrightarrow k^2=\left(\pm2\right)^2\\ \Rightarrow k\in\left\{2;-2\right\}\)
+Khi \(k=2\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
+Khi \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-2.3=-6\\y=-2.7=-14\end{matrix}\right.\)
Vậy...
Ta có: \(\frac{3}{x}=\frac{y}{7}.\)
\(\Rightarrow\frac{x}{3}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Có: \(x.y=84\)
=> \(3k.7k=84\)
=> \(21k^2=84\)
=> \(k^2=84:21\)
=> \(k^2=4\)
=> \(k=\pm2.\)
TH1: \(k=2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
TH2: \(k=-2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right).3=-6\\y=\left(-2\right).7=-14\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;14\right),\left(-6;-14\right).\)
Chúc bạn học tốt!
