\(\hept{\begin{cases}\left(x+1\right)+\sqrt{x}+\sqrt{y+1}=2\\\left(y+1\right)+\sqrt{y}+\sqrt{x+1}=2\end{cases}}\) ĐK: \(\hept{\begin{cases}x\ge0\\y\ge0\end{cases}}\)
Lấy pt (1) - (2) Ta được
\(\left(x+1\right)-\left(y+1\right)+\sqrt{x}-\sqrt{y}+\left(\sqrt{y+1}-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left(x-y\right)+\left(\sqrt{x}-\sqrt{y}\right)+\frac{\left(y+1\right)-\left(x+1\right)}{\sqrt{y+1}+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y+1}+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1-\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y+1}+\sqrt{x+1}}\right)=0\)
