\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-4x+1+4y^2-8y+4=22\\\left(x^2-x\right)\left(y^2-2y\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-x\right)+1+4\left(y^2-2y\right)+4=22\\\left(x^2-x\right)\left(y^2-2y\right)=1\end{matrix}\right.\)
Đặt \(x^2-x=a;y^2-2y=b\)
\(\Rightarrow\left\{{}\begin{matrix}4a+4b=17\\ab=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a+\frac{4}{a}=17\\ab=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=4\\a=\frac{1}{4}\end{matrix}\right.\\ab=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=\frac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\frac{1}{4}\\b=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x=4\\y^2-2y=\frac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x=\frac{1}{4}\\y^2-2y=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1+\sqrt{17}}{2}\\x=\frac{1-\sqrt{17}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}y=\frac{2+\sqrt{5}}{2}\\y=\frac{2-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1+\sqrt{2}}{2}\\x=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}y=1+\sqrt{5}\\y=1-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
P/s: cái bài này lắm nghiệm quá, bn tự kl nhé, mk lười :))