chết nhầm cho sửa lại
Đặt \(\frac{x}{3}=-\frac{y}{7}=k\)
\(\Rightarrow\frac{x}{3}=3k;-\frac{y}{7}=-7k\)
Theo đề bài ra , ta có :
\(3k.-7k=-189\)
\(\Leftrightarrow-21k^2=-189\)
\(\Leftrightarrow k^2=9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}k=3\\k=-3\end{array}\right.\)
Khi \(k=3\) , thì :
\(\left[\begin{array}{nghiempt}x=6\\y=-21\end{array}\right.\)
Khi \(k=-3\) , thì :
\(\left[\begin{array}{nghiempt}x=-6\\y=21\end{array}\right.\)
Vậy ................
Đặt \(\frac{x}{3}=-\frac{y}{7}=k\)
\(\Rightarrow\frac{x}{3}=3k;-\frac{y}{7}=-7k\)
Theo đề bài ta có :
\(3k.-7k=-189\)
\(\Leftrightarrow-21k^2=-189\)
\(\Leftrightarrow k^2=9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}k=9\\k=-9\end{array}\right.\)
Khi \(k=9\) , thì :
\(\left[\begin{array}{nghiempt}x=27\\y=-63\end{array}\right.\)
Khi \(k=-9\) , thì :
\(\left[\begin{array}{nghiempt}x=-27\\x=63\end{array}\right.\)
Vậy .................