$(\sqrt{x+1}-\sqrt{y-2})^{2}=x+1+y-2-2\sqrt{(x+1)(y-2)}=2a+2-1-2\sqrt{(x+1)(y-2)}=2a+1-2\sqrt{(x+1)(y-2)}=a^{2}\Rightarrow a^{2}-2a-1\leq 0\Rightarrow (a-1)^{2}\leq 2\Rightarrow -\sqrt{2}\leq a-1\leq \sqrt{2}\Rightarrow 1-\sqrt{2}\leq a\leq \sqrt{2}+1$
đáp án là $0\leq a\leq 2$ bạn à