Question

giải hệ phương trình :\(\left\{{}\begin{matrix}x+y-2xy=0\\x+y-x^2y^2=\sqrt{\left(xy-1\right)^2+1}\end{matrix}\right.\)

Answer 1

Đặt \(\left\{{}\begin{matrix}x+y=s\\xy=p\end{matrix}\right.\) \(\left(s^2\ge4p\right)\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}s-2p=0\left(1\right)\\s-p^2=\sqrt{\left(p-1\right)^2+1}\left(2\right)\end{matrix}\right.\)

Trừ (2) cho (1) \(\Rightarrow2p-p^2=\sqrt{\left(p-1\right)^2+1}\)

\(\Rightarrow\left\{{}\begin{matrix}0\le p\le2\\p^4-4p^3+4p^2=p^2-2p+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0\le p\le2\\p=1\left(tm\right),p=1\pm\sqrt{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\)Đến đây bạn thay vào tự giải.