Question

giải hệ: \(\left\{{}\begin{matrix}x+y=2\\2x^2+y+xy=8\end{matrix}\right.\)

Answer 1

Lời giải:

HPT \(\Leftrightarrow \)\(\left\{\begin{matrix} x+y=2\\ x^2+(x^2+xy)+y=8\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y=2\\ x^2+x(x+y)+y=8\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x+y=2\\ x^2+2x+y=8\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=8\)

\(\Leftrightarrow x^2+x-6=0\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

Nếu \(x=2\Rightarrow y=2-x=0\). Ta có cặp \((x,y)=(2,0)\) thỏa mãn

Nếu \(x=-3\Rightarrow y=2-x=5\). Ta có cặp \((x,y)=(-3,5)\) thỏa mãn.