Điều kiện của $x,y$ là gì vậy bạn? Nếu chỉ có 1 PT như thế kia thì không tìm được $x,y$ cụ thể bạn nhé.
\(x^2-2x+1-4y^2=5\)
=>\(\left(x-1\right)^2-4y^2=5\)
=>\(\left(x-1-2y\right)\left(x-1+2y\right)=5\)
=>\(\left(x-2y-1\right)\left(x+2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
TH1: \(\left\{{}\begin{matrix}x-2y-1=1\\x+2y-1=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=2\\x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\2y=x-2=4-2=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x-2y-1=5\\x+2y-1=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=6\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\x+2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\2y=2-x=2-4=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x-2y-1=-1\\x+2y-1=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=0\\x+2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\x=2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2\\2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x-2y-1=-5\\x+2y-1=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=-4\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\x=-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2\\-2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)
