$6x^{2}-3xy+x+y=1\Leftrightarrow 6x^2+x-1-y(3x-1)=0\Leftrightarrow (3x-1)(2x+1-y)=0$
$\Leftrightarrow \left[ {\begin{matrix} x=1/3\\ y=2x+1 \end{matrix}} \right.$
+ Nếu $x=1/3$. Ta có
$1+y+\sqrt{1+y^{2}}=2\Leftrightarrow \sqrt{1+y^{2}}=1-y\Leftrightarrow \begin{cases}y \le 1 \\ 1+y^2=(1-y)^2 \end{cases}\Leftrightarrow y=0.$
+ Nếu $y=2x+1$. Ta có
$5x+1+\sqrt{4x^2+7x+1}=2\Leftrightarrow \sqrt{4x^2+7x+1}=1-5x\Leftrightarrow \begin{cases}x \le 1/5 \\ 4x^2+7x+1= (1-5x)^2 \end{cases}\Leftrightarrow x=0\Leftrightarrow y=1.$