\(\left\{{}\begin{matrix}x^2+y^2+3=4x\\x^3+12x+y^3=6x^2+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-4x+4\right)+y^2=1\\\left(x^3-6x^2+12x-8\right)+y^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+y^2=1\\\left(x-2\right)^3+y^3=1\end{matrix}\right.\)
Đặt \(a=x-2;b=y\). Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}a^2+b^2=1\\a^3+b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-1\\\left(a+b\right)\left(a^2+b^2-ab\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-1\\\left(a+b\right)\left(1-\dfrac{\left(a+b\right)^2-1}{2}\right)=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(a+b\right)\left[3-\left(a+b\right)^2\right]=2\)
\(\Leftrightarrow3\left(a+b\right)-\left(a+b\right)^3=2\)
\(\Leftrightarrow\left(a+b\right)^3-3\left(a+b\right)+2=0\)
\(\Leftrightarrow\left(a+b\right)^3-\left(a+b\right)^2+\left(a+b\right)^2-\left(a+b\right)-2\left(a+b-1\right)=0\)
\(\Leftrightarrow\left(a+b\right)^2\left(a+b-1\right)+\left(a+b\right)\left(a+b-1\right)-2\left(a+b-1\right)=0\)
\(\Leftrightarrow\left(a+b-1\right)\left[\left(a+b\right)^2+\left(a+b\right)-2\right]=0\)
\(\Leftrightarrow\left(a+b-1\right)^2\left(a+b+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=1\\a+b=-2\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}a+b=1\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\left(a+b\right)^2-2ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\ab=0\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(0;1\right),\left(1;0\right)\)
\(\Rightarrow\left(x-2;y\right)=\left(0;1\right),\left(1;0\right)\)
\(\Rightarrow\left(x;y\right)=\left(2;1\right),\left(3;0\right)\)
Với \(\left\{{}\begin{matrix}a+b=-2\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=-2\\\left(a+b\right)^2-2ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=-2=S\\ab=\dfrac{3}{2}=P\end{matrix}\right.\left(2\right)\)
Ta có: \(S^2-4P=\left(-2\right)^2-4.\dfrac{3}{2}=-2< 0\)
\(\Rightarrow\)Không tồn tại số a,b nào thỏa hệ phương trình (2).
Vậy nghiệm (x;y) của hpt đã cho là \(\left(2;1\right),\left(3;0\right)\)
