Cách 1:
Điều kiện: $x,y\ge0$
Ta có:
$\left\{\begin{array}{l}(\sqrt{x+1}+\sqrt{x})+(\sqrt{y+1}+\sqrt{y})=2\\(\sqrt{x+1}-\sqrt{x})-(\sqrt{y+1}-\sqrt{y})=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(\sqrt{x+1}+\sqrt{x})+(\sqrt{y+1}+\sqrt{y})=2\\\frac{1}{\sqrt{x+1}+\sqrt{x}}-\frac{1}{\sqrt{y+1}+\sqrt{y}}=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(\sqrt{x+1}+\sqrt{x})+(\sqrt{y+1}+\sqrt{y})=2\\\sqrt{x+1}+\sqrt{x}=\sqrt{y+1}+\sqrt{y}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sqrt{x+1}+\sqrt{x}=1\\\sqrt{y+1}+\sqrt{y}=1\end{array}\right.$
Xét hàm: $f(t)=\sqrt{t+1}+\sqrt{t}, t\ge0$
Ta có: $f'(t)=\frac{1}{2\sqrt{t+1}}+\frac{1}{2\sqrt t}>0, \forall t>0$.
Suy ra $f(t)=1$ có nhiều nhất 1 nghiệm.
Mà $f(0)=1 \Rightarrow (x,y)=(0,0)$