ĐK : $y\geq 1;6\geq x\geq 0$
$xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}$$\Leftrightarrow x(y-1)-(y-1)^2=\sqrt{y-1}-\sqrt{x}$
$\Leftrightarrow (y-1)(y-1-x)+\sqrt{y-1}-\sqrt{x}=0$
$\Leftrightarrow (\sqrt{y-1}-\sqrt{x})((y-1)(\sqrt{y-1}+\sqrt{x})+1)=0$
$\Leftrightarrow \sqrt{y-1}=\sqrt{x}$
$\Leftrightarrow x=y-1$ Thay vô Pt sau rồi tự giải tiếp ...