Nhân 2 pt đầu r cộng pt mới cho pt 2 xong giải
\(\left\{{}\begin{matrix}2x^2+y^2-4x+2y=1\\3x^2-2y^2-6x-4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x^2+2y^2-8x+4y=2\left(1\right)\\3x^2-2y^2-6x-4y=5\left(2\right)\end{matrix}\right.\)
Lấy (1) cộng (2) ta được: \(7x^2-14x=7\Leftrightarrow x^2-2x=1\Leftrightarrow x^2-2x-1=0\Leftrightarrow\left\{{}\begin{matrix}x_1=1+\sqrt{2}\\x_2=1-\sqrt{2}\end{matrix}\right.\)
+) T/h 1: \(x=1+\sqrt{2}\) thay vào (1) ta có:
\(\left(1\right)\Leftrightarrow2\left(1+\sqrt{2}\right)^2+y^2-4\left(1+\sqrt{2}\right)+2y=1\\ \Leftrightarrow2\left(1+2\sqrt{2}+2\right)-4\left(1+\sqrt{2}\right)+y^2+2y-1=0\\ \Leftrightarrow2+4\sqrt{2}+4-4-4\sqrt{2}+y^2+2y-1=0\\ \Leftrightarrow y^2+2y+1=0\\ \Leftrightarrow\left(y+1\right)^2=0\Leftrightarrow y=-1\)
+) T/h 2 : \(x=1-\sqrt{2}\) thay vào (1) ta có:
\(\left(1\right)\Leftrightarrow2\left(1-\sqrt{2}\right)^2+y^2-4\left(1-\sqrt{2}\right)+2y=1\\ \Leftrightarrow2\left(1-2\sqrt{2}+2\right)-4\left(1-\sqrt{2}\right)+y^2+2y-1=0\\ \Leftrightarrow2-4\sqrt{2}+4-4+4\sqrt{2}+y^2+2y-1=0\\ \Leftrightarrow y^2+2y+1=0\\ \Leftrightarrow\left(y+1\right)^2=0\Leftrightarrow y=-1\)
Vậy hệ phương trình có nghiệm là: \(\left(x;y\right)=\left(1+\sqrt{2};-1\right);\left(1-\sqrt{2};-1\right)\)