Đặt \(\left\{{}\begin{matrix}y^2-4y=a\\2y-x=b\end{matrix}\right.\)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\a+b=3\end{matrix}\right.\)
\(\Leftrightarrow a\left(3-a\right)=2\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\Leftrightarrow b=1\\a=1\Leftrightarrow b=2\end{matrix}\right.\)
TH1: a = 2; b = 1
Có: \(y^2-4y=2\)
\(\Leftrightarrow\left(y-2\right)^2=6\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}y=\sqrt{6}+2\\y=-\sqrt{6}+2\end{matrix}\right.\)
- Nếu \(y=\sqrt{6}+2\):
Có: \(2y-x=1\)
\(\Leftrightarrow2\sqrt{6}+4-x=1\)
\(\Leftrightarrow x=2\sqrt{6}+3\)
- Nếu \(y=-\sqrt{6}+2\)
Có: \(2y-x=1\)
\(\Leftrightarrow-2\sqrt{6}+4-x=1\)
\(\Leftrightarrow x=-2\sqrt{6}+3\)
TH2: Nếu a = 1; b = 2
Có: \(y^2-4y=1\)
\(\Leftrightarrow\left(y-2\right)^2=5\)
\(\Leftrightarrow\left[{}\begin{matrix}y=\sqrt{5}+2\\y=-\sqrt{5}+2\end{matrix}\right.\)
- Nếu \(y=\sqrt{5}+2\)
Có: \(2y-x=2\)
\(\Leftrightarrow2\sqrt{5}+4-x=2\)
\(\Leftrightarrow x=2\sqrt{5}+2\)
- Nếu \(y=-\sqrt{5}+2\)
Có: \(2y-x=2\)
\(\Leftrightarrow-2\sqrt{5}+4-x=2\)
\(\Leftrightarrow x=-2\sqrt{5}+2\)
Vậy HPT có nghiệm \(\left(x;y\right)=\left\{\left(2\sqrt{6}+3;\sqrt{6}+2\right);\left(-2\sqrt{6}+3;-\sqrt{6}+2\right);\left(2\sqrt{5}+2;\sqrt{5}+2\right);\left(-2\sqrt{5}+2;-\sqrt{5}+2\right)\right\}\)
