\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(y-3\right)=9\\\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{3}{2}\right)^2=\dfrac{81}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=u\\y-\dfrac{3}{2}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(u-\dfrac{3}{2}\right)\left(v-\dfrac{3}{2}\right)=9\\u^2+v^2=\dfrac{81}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-\dfrac{3}{2}\left(u+v\right)=\dfrac{27}{4}\\\left(u+v\right)^2-2uv=\dfrac{81}{2}\end{matrix}\right.\) \(\Rightarrow\left(u+v\right)^2-3\left(u+v\right)-54=0\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=9\Rightarrow uv=\dfrac{81}{4}\\u+v=-6\Rightarrow uv=\dfrac{-9}{4}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}u+v=9\\uv=\dfrac{81}{4}\end{matrix}\right.\) \(\Rightarrow u^2-9u+\dfrac{81}{4}=0\Rightarrow u=v=\dfrac{9}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=u+\dfrac{1}{2}=5\\y=v+\dfrac{3}{2}=6\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}u+v=-6\\uv=\dfrac{-9}{4}\end{matrix}\right.\)
\(\Rightarrow u^2+6u-\dfrac{9}{4}=0\Rightarrow\left[{}\begin{matrix}u=\dfrac{-6+3\sqrt{5}}{2}\Rightarrow v=\dfrac{-6-3\sqrt{5}}{2}\\u=\dfrac{-6-3\sqrt{5}}{2}\Rightarrow v=\dfrac{-6+3\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5+3\sqrt{5}}{2}\\y=\dfrac{-3-3\sqrt{5}}{2}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=\dfrac{-5-3\sqrt{5}}{2}\\y=\dfrac{-3+3\sqrt{5}}{2}\end{matrix}\right.\)
