Câu 2.Hệ $\Leftrightarrow \begin{cases}(x-1)^2-(y+1)=0 \\ (x+y)^2-y(x-y)-2xy-1=0 \end{cases}$ $(\bigstar)$
Đặt $\begin{cases}a=x-1 \\ b=y+1 \end{cases}\Rightarrow \begin{cases}ab=(x-1)(y+1)=xy+x-y-1 \\ a+b=x+y \\a-b+2=x-y \end{cases}$
$\Rightarrow ab=xy+a-b+1\Rightarrow xy=ab-a+b-1$
Hệ $(\bigstar)\Leftrightarrow \begin{cases}a^2=b \\ (a+b)^2-(b-1)(a-b+2)-2(ab-a+b-1)-1=0 \end{cases}$
$\Rightarrow a^2(a+1)^2+(a+1)^2(a-1)(a-2)-2(a+1)^2(a-1)-1=0$
$\Leftrightarrow (a+1)^2(2a^2-5a+4)=1$