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\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=y\left(x-2\right)x\left(y-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=\left(x^2-2x\right)\left(y^2-4y\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)
\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)
\(\left\{{}\begin{matrix}x^3+xy^2+3\left(x-2y\right)=0\\x^2+xy=3\end{matrix}\right.\)\(\Rightarrow x^3+xy^2+\left(x^2+xy\right)\left(x-2y\right)=0\)\(\Leftrightarrow x^3+xy^2+x^3-x^2y-2xy^2=0\Leftrightarrow2x^3-x^2y-xy^2=0\)\(\Leftrightarrow x\left(2x+y\right)\left(x-y\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=-2x\\x=y\end{matrix}\right.\)
+) \(x=0\Rightarrow0y=3\)(vô nghiệm)
+) y=-2x \(\Rightarrow x^2-2x^2=3\Leftrightarrow-x^2=3\)(vô nghiệm)
+) x=y\(\Rightarrow2x^2=3\Leftrightarrow x^2=\dfrac{3}{2}\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{\dfrac{3}{2}}\\x=y=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)
\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)
Thế x=1 vào 1 trong 2 phương trình => y=1
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+x=u\\y^2+y=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;2\right);\left(2;6\right)\)
TH1: \(\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\) \(\Rightarrow...\)
TH2: ... tương tự
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\\left(x+y\right)xy=30\end{matrix}\right.\)
Theo Viet đảo, \(x+y\) và \(xy\) là nghiệm của:
\(t^2-11t+30=0\Rightarrow\left[{}\begin{matrix}t=6\\t=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=6\\xy=5\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=5\\xy=6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;5\right);\left(5;1\right);\left(2;3\right);\left(3;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
Đặt \(x+y=S;xy=P\Rightarrow\left\{{}\begin{matrix}P+S=11\\P.S=30\end{matrix}\right.\)
\(\Rightarrow\frac{30}{S}+S=11\Leftrightarrow30+S^2=11S\Leftrightarrow\left[{}\begin{matrix}S=6\\S=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}P=5\\P=6\end{matrix}\right.\)
Xét \(\left\{{}\begin{matrix}P=6\\S=5\end{matrix}\right.\Rightarrow X^2-5X+6=0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\end{matrix}\right.\)
Làm tương tự vs trường hợp còn lại