\(\left\{{}\begin{matrix}\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}-4\sqrt{y}=-4\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=1\\\sqrt{x}-2\sqrt{y}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}-2\sqrt{1}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}-2=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\) vậy \(y=1;x=0\)
\(\left\{{}\begin{matrix}\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\)
Đặt \(u=\sqrt{x};v=\sqrt{y}\) cho đơn giản (\(u;v\ge0\))
\(\left\{{}\begin{matrix}u-2v=-2\\2u-3v=-3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2u-4v=-4\\2u-3v=-3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-v=-1\\u-2v=-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}v=1\\u-2.1=-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}v=1\\u=0\end{matrix}\right.\left(TMĐK\right)\)
\(\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)
