=> \(\hept{\begin{cases}x^2+2xy+y^2-4x+4y=12\\x^2-2xy+y^2-2x-2y=3\end{cases}}\)
Rồi đến đây tự làm nhé
HPT <=> \(\hept{\begin{cases}\left(x+y\right)^2-4\left(x+y\right)+4=16\\\left(x-y\right)^2-2\left(x-y\right)+1=4\end{cases}}\)<=> \(\hept{\begin{cases}\left(x+y-2\right)^2=4^2\\\left(x-y-1\right)^2=2^2\end{cases}}\)
=> \(\hept{\begin{cases}x+y-2=\pm4\\x-y-1=\pm2\end{cases}}\)
Có các TH:
1/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{9}{2}\\y=\frac{3}{2}\end{cases}}\)
2/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{7}{2}\end{cases}}\)
3/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{5}{2}\end{cases}}\)
4/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{1}{2}\end{cases}}\)
