Ta có:
\(\left\{{}\begin{matrix}2x+y=3\\x^2+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-x^2=-2\\2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+2x+2=0\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\left(x^2-2x+1\right)+3=0\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\left(x-1\right)^2=-3\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=3\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{3}+1\\x=1-\sqrt{3}\end{matrix}\right.\\y=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{3}+1\\y=1-2\sqrt{3}\end{matrix}\right.\\\left[{}\begin{matrix}x=1-\sqrt{3}\\y=1+2\sqrt{3}\end{matrix}\right.\end{matrix}\right.\)
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