\(\left\{{}\begin{matrix}3x^2+xy-4x+2y=2\\x\left(x+1\right)+y\left(y+1\right)=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x^2+xy-4x+2y=2\left(1\right)\\x^2+x+y^2+y=4\left(2\right)\end{matrix}\right.\)
Trừ vế theo vế của (1) cho (2)\(\Leftrightarrow3x^2+xy-4x+2y-x^2-x-y^2-y=-2\Leftrightarrow2x^2-y^2-5x+y+xy+2=0\Leftrightarrow\left(2x-y-1\right)\left(x+y-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=2x-1\\y=2-x\end{matrix}\right.\)
TH1: y=2x-1
thay vào (2)\(\Leftrightarrow x^2+x+\left(2x-1\right)^2+2x-1=4\Leftrightarrow x^2+x+4x^2-4x+1+2x-5=0\Leftrightarrow5x^2-x-4=0\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-\frac{4}{5}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=1\\y=-\frac{8}{5}\end{matrix}\right.\)
TH2: y=2-x
Thay vào (2)\(\Leftrightarrow x^2+x+\left(2-x\right)^2+2-x=4\Leftrightarrow x^2+x+4-4x+x^2+2-x=4\Leftrightarrow2x^2-4x+2=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)\(\Leftrightarrow y=1\)
Vậy (x;y)={(1;1);(\(-\frac{4}{5};-\frac{8}{5}\))}
