c/ \(\dfrac{x}{y}=-\dfrac{2}{3}\Leftrightarrow\dfrac{x}{-2}=\dfrac{y}{3}\)
Đặt : \(\dfrac{x}{-2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=-2k\\y=3k\end{matrix}\right.\)
Lại có : \(xy=96\)
\(\Leftrightarrow-6k^2=96\Leftrightarrow k^2=-16\) (vô lí)
d/ \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=k\Leftrightarrow\left\{{}\begin{matrix}x^2=9k\\y^2=16k\end{matrix}\right.\)
\(x^2+y^2=100\)
\(\Leftrightarrow25k=100\Leftrightarrow k=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\end{matrix}\right.\)
Vậy..
d:Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=4\)
Do đó: \(\left\{{}\begin{matrix}x^2=36\\y^2=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;-6\right\}\\y\in\left\{8;-8\right\}\end{matrix}\right.\)
