ĐK: \(x,y\ge0\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\left(1\right)\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\)
Cộng vế theo vế 2 phương trình ta được: \(7\sqrt{x}=0\Leftrightarrow x=0\)
Khi đó \(\left(1\right)\Leftrightarrow-2\sqrt{y}=-2\Leftrightarrow y=1\)
Vậy hệ đã cho có nghiệm \(\left(x;y\right)=\left(0;1\right)\)
ĐKXĐ: \(x\ge0;y\ge0\)
\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x}=0\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2\sqrt{0}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\0+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\) (TM)
Vậy...
\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}7\sqrt{x}=0\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\sqrt{x}=0\\2.0+\sqrt{y}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=0\\\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=0\\y=^+_11\end{matrix}\right.\)
