\(đặt\) \(\sqrt{2x-1}=t\left(t\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}t-y\left(1+2t\right)=-8\\-y^2-yt=t^2-12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t-y-2yt=-8\\-y^2-yt-t^2=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t-y-2yt=-8\\t^2-2yt+y^2+3yt=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t-y-2yt=-8\\\left(t-y\right)^2+3yt=12\end{matrix}\right.\)
\(đặt\) \(\left\{{}\begin{matrix}t-y=a\\yt=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-2b=-8\\a^2+3b=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b=4\\b=\dfrac{13}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\a=-\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow TH1:\left\{{}\begin{matrix}t-y=0\\yt=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t=y=-2\left(ktm\right)\\t=y=2\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}t=2=\sqrt{2x-1}\\y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2,5\\y=2\end{matrix}\right.\)
\(\Rightarrow TH2:\left\{{}\begin{matrix}t-y=\dfrac{-3}{2}\\yt=\dfrac{13}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t=\dfrac{1}{4}\left(-3-\sqrt[]{61}\right),y=\dfrac{1}{4}\left(3-\sqrt{61}\right)\left(ktm\right)\\t=\dfrac{1}{4}\left(\sqrt{61}-3\right);y=\dfrac{1}{4}\left(3+\sqrt{61}\right)\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{2x-1}=\dfrac{1}{4}\left(\sqrt{61}-3\right)\\y=\dfrac{1}{4}\left(3+\sqrt{61}\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{16}\left(43-3\sqrt{61}\right)\\y=\dfrac{1}{4}\left(3+\sqrt{61}\right)\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left\{...\right\}\)
