\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\xy=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3x=2y\\xy-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\xy-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3xy-2y^2=0\\3xy-18=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}-2y^2-\left(-18\right)=0\\3xy-2y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\9x-18=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
d: Đặt \(\dfrac{x}{7}=\dfrac{y}{-2}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7k\\y=-2k\end{matrix}\right.\)
Ta có: \(x^2y=-98\)
\(\Leftrightarrow-98k^3=-98\)
hay k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=7k=7\\y=-2k=-2\end{matrix}\right.\)
cái bài n mk biết làm mà do muốn rõ hơn nên các bn làm chi tiết hộ mk nha ! Cảm ơn nhiều !
